1001:排序完按照题意做即可。
1 #include
2 #include
3 #include
4 #include
5 using namespace std;
6 int a[10001],k,n;
7 int main(){
8 scanf("%d%d",&n,&k);
9 for (int i=1; i<=n; i++) scanf("%d",&a[i]);
10 sort(a+1,a+n+1);
11 int m=a[n-k+1]-a[k];
12 if (m<2&&m>=0) printf("NO");
13 else if (m>0){
14 if (m==2) printf("YES\n");
15 else{
16 for (int i=2; i<=floor(sqrt(m)); i++)
17 if (m%i==0){
18 printf("NO\n");
19 printf("%d",m);
20 return 0;
21 }
22 printf("YES\n");
23 }
24 printf("%d",m);
25 return 0;
26 }else printf("NO\n%d",m);
27 return 0;
28 }
1001
1002：模拟，按题意完成即可。
1 #include
2 #include
3 #include
4 using namespace std;
5 int qm,bj,lw,hv,ma,n,sum;
6 bool gb,wt;
7 string bb,cc;
8 int main(){
9 cin>>n;
10 for (int i=1; i<=n; i++){
11 hv=0;
12 //cout< 13 cin>>bb; 14 scanf("%d%d",&qm,&bj); 15 getchar(); 16 gb=getchar()=='Y'?1:0; 17 getchar(); 18 wt=getchar()=='Y'?1:0; 19 scanf("%d",&lw); 20 if (qm>80&&lw) hv+=8000; 21 if (qm>85&&bj>80) hv+=4000; 22 if (qm>90) hv+=2000; 23 if (qm>85&&wt) hv+=1000; 24 if (bj>80&&gb) hv+=850; 25 if (hv>ma) { 26 ma=hv; 27 cc=bb; 28 } 29 sum+=hv; 30 //cout< 31 } 32 cout< 33 } 1002 1003：模拟即可。 #include #include #include #include using namespace std; int main(){ int m,t,u,f,d,s=0,t1=0; char a[100001]; scanf("%d %d %d %d %d",&m,&t,&u,&f,&d); for(int i=0;i scanf("%s",&a[i]); for(int i=0;i if(a[i]=='u'||a[i]=='d') s=s+u+d; else s=s+f*2; if(s>m) break; t1++; } cout< return 0; } 1003 1004：用了奇技淫巧。。。正解应该是记忆化搜索，我的做法是排序后贪心一波。 1 #include 2 #include 3 #include 4 using namespace std; 5 struct fff{ 6 int x,y,n; 7 }a[10001]; 8 int f[101][101],b[101][101],n,m,ans; 9 int pd(const fff &a,const fff &b){ 10 return a.n 11 } 12 int main(){ 13 cin>>n>>m; 14 for (int i=1; i<=n; i++) 15 for (int j=1; j<=m; j++){ 16 scanf("%d",&b[i][j]); 17 a[i*m-m+j].x=i;a[i*m-m+j].y=j;a[i*m-m+j].n=b[i][j]; 18 f[i][j]=1; 19 } 20 sort(a+1,a+n*m+1,pd); 21 for (int i=1; i<=n*m; i++){ 22 int x=a[i].x,y=a[i].y; 23 f[x+1][y]=((b[x+1][y]>a[i].n)&&(f[x+1][y] 24 f[x-1][y]=((b[x-1][y]>a[i].n)&&(f[x-1][y] 25 f[x][y+1]=((b[x][y+1]>a[i].n)&&(f[x][y+1] 26 f[x][y-1]=((b[x][y-1]>a[i].n)&&(f[x][y-1] 27 } 28 for (int i=1; i<=n; i++) 29 for (int j=1; j<=m; j++) 30 ans=max(ans,f[i][j]); 31 cout< 32 } 1004 1005：裸01背包。。。。 1 #include 2 #include 3 using namespace std; 4 int f[1001],p[101],a[101],n,v; 5 int main(){ 6 cin>>v>>n; 7 for (int i=1; i<=n; i++) scanf("%d%d",&a[i],&p[i]); 8 for (int i=1; i<=n; i++) 9 for (int j=v; j>=a[i]; j--) 10 f[j]=max(f[j],p[i]+f[j-a[i]]); 11 cout< 12 } 1005 1006：模拟。 #include #include #define k(x,y) y*(a[x]-48) using namespace std; int main(){ char a[14]; gets(a); if ((k(0,1)+k(2,2)+k(3,3)+k(4,4)+k(6,5)+k(7,6)+k(8,7)+k(9,8)+k(10,9))%11==(a[12]=='X'?10:a[12]-48)) cout<<"Right"; else{ cout< if ((k(0,1)+k(2,2)+k(3,3)+k(4,4)+k(6,5)+k(7,6)+k(8,7)+k(9,8)+k(10,9))%11==10) cout<<"X"; else cout<<(k(0,1)+k(2,2)+k(3,3)+k(4,4)+k(6,5)+k(7,6)+k(8,7)+k(9,8)+k(10,9))%11; } } 1006 1007：各种排序。。。然后就行了 1 #include 2 #include 3 #include 4 using namespace std; 5 int m,n,k,l,d,x,xx,yy,y; 6 struct fff{ 7 int k,no; 8 }anh[1001],anl[1001]; 9 int fuck(const fff &a,const fff &b){ 10 return a.k>b.k; 11 } 12 int fuck233(const fff &a,const fff &b){ 13 return a.no 14 } 15 int main(){ 16 cin>>m>>n>>k>>l>>d; 17 for (int i=1; i 18 for (int i=1; i 19 for (int i=1; i<=d; i++){ 20 scanf("%d%d%d%d",&x,&y,&xx,&yy); 21 if (x==xx) anl[min(y,yy)].k++; 22 else if (y==yy) anh[min(x,xx)].k++; 23 } 24 sort(anl+1,anl+n+1,fuck); 25 sort(anh+1,anh+1+m,fuck); 26 sort(anl+1,anl+l+1,fuck233); 27 sort(anh+1,anh+k+1,fuck233); 28 for (int i=1; i 29 cout< 30 for (int i=1; i 31 cout< 32 } 1007 1008：又是模拟。。。 1 #include 2 #include 3 using namespace std; 4 int f[31][31],m,n; 5 int main(){ 6 scanf("%d%d",&n,&m); 7 f[0][1]=1; 8 for (int i=1; i<=m; i++) 9 for (int j=1; j<=n; j++){ 10 f[i][j]=f[i-1][j==1?n:j-1]+f[i-1][j==n?1:j+1]; 11 } 12 cout< 13 } 1008 1009：略过略过~ 1010：模拟即可。。。 1 #include 2 #include 3 #include 4 #include 5 using namespace std; 6 char a[100]; 7 int f[27],ma,mi=0x7f; 8 int main(){ 9 gets(a); 10 for (int i=0; i 11 f[a[i]-95]++; 12 for (int i=1; i<=26; i++){ 13 ma=max(ma,f[i]); 14 mi=min(mi,f[i]==0?0x7f:f[i]); 15 } 16 ma-=mi; 17 if (ma<2){ 18 cout<<"No Answer\n0"; 19 return 0; 20 } 21 else{ 22 if (ma!=2) for (int i=2; i<=floor(sqrt(ma)); i++) if (ma%i==0){ 23 cout<<"No Answer\n0"; 24 return 0; 25 } 26 cout<<"Lucky Word\n"< 27 } 28 } 1010 1011：题意就是从(1,1)->(m,n)找两条最大权值不相交路线。f[i,j]就是找到了第i个人，第一条路线在第j行，第二条路线在第k行的最优值。DP一下就行了。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 using namespace std; 17 int f[101][51][51],a[51][51],n,m; 18 inline int in(){ 19 int x=0,f=1; 20 char ch=getchar(); 21 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 22 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 23 return x*f; 24 } 25 int main(){ 26 n=in(),m=in(); 27 For(i,1,n) 28 For(j,1,m) 29 a[i][j]=in(); 30 For(i,1,n+m) 31 For(j,1,min(n,i)) 32 For(k,1,min(n,i)) 33 if (j!=k||i==n+m) 34 f[i][j][k]=max(f[i-1][j-1][k-1],max(max(f[i-1][j][k],f[i][j][k]),max(f[i-1][j-1][k],f[i-1][j][k-1])))+a[j][i-j+1]+a[k][i-k+1]; 35 printf("%d\n",f[n+m][n][n]); 36 } 1011 1012：按题意枚举之后判断就行了。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 int n,a[10]={6,2,5,5,4,5,6,3,7,6}; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 inline int get(int x){ 28 if (!x) return 6; 29 int cnt=0; 30 while(x){ 31 cnt+=a[x%10]; 32 x/=10; 33 } 34 return cnt; 35 } 36 int main(){ 37 n=in(); 38 n-=4; 39 int ans=0; 40 For(i,0,800) 41 For(j,i,800) 42 if (get(i)+get(j)+get(i+j)==n) 43 if (i!=j) ans+=2; 44 else ans++; 45 printf("%d",ans); 46 } 1012 1013:二维费用的01背包。我们用f[i][j]表示剩余i金j金币时能跑到zxy妹子的最多个数，记录此时的最小时间消耗为g[i][j](你泡妹子还那么急真是不懂得享受┑(￣Д ￣)┍)状态转移方程略。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 struct zxy{ 20 int rp,c,t; 21 }a[101]; 22 int f[101][101],g[101][101],n,rp,m,ans,mi; 23 inline int in(){ 24 int x=0,f=1; 25 char ch=getchar(); 26 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 27 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 28 return x*f; 29 } 30 int main(){ 31 n=in(); 32 For(i,1,n) 33 a[i].c=in(),a[i].rp=in(),a[i].t=in(); 34 m=in(),rp=in(); 35 mem(g,127/3); g[0][0]=0; 36 For(i,1,n) 37 Ford(j,m,a[i].c) 38 Ford(k,rp,a[i].rp) 39 if (f[j-a[i].c][k-a[i].rp]+1>f[j][k]||(f[j-a[i].c][k-a[i].rp]+1==f[j][k]&&g[j][k]>g[j-a[i].c][k-a[i].rp]+a[i].t)){ 40 f[j][k]=f[j-a[i].c][k-a[i].rp]+1; 41 g[j][k]=g[j-a[i].c][k-a[i].rp]+a[i].t; 42 if (f[j][k]>ans||(f[j][k]==ans&&g[j][k] 43 ans=f[j][k]; 44 mi=g[j][k]; 45 } 46 } 47 printf("%d",mi); 48 } 1013 1014:区间DP，枚举区间内要拿走的牌按题意完成即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 int n,a[101],f[101][101]; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 inline int dp(int l,int r){ 28 if (f[l][r]!=f[0][0]) return f[l][r]; 29 if (l+1==r) return 0; 30 For(i,l+1,r-1) 31 f[l][r]=min(f[l][r],dp(l,i)+dp(i,r)+a[l]*a[i]*a[r]); 32 return f[l][r]; 33 } 34 int main(){ 35 n=in(); 36 For(i,1,n) a[i]=in(); 37 mem(f,127/3); 38 dp(1,n); 39 printf("%d",f[1][n]); 40 } 1014 1015：非常水的DP。 1 #include 2 #include 3 #include 4 using namespace std; 5 int a[111],n,f[11]; 6 int main(){ 7 memset(a,127/3,sizeof(a)); 8 for (int i=1; i<=10; i++) 9 scanf("%d",&f[i]); 10 cin>>n; 11 a[10]=0; 12 for (int i=11; i<=n+10; i++) 13 for (int j=1; j<=10; j++) 14 a[i]=min(a[i],a[i-j]+f[j]); 15 cout< 16 } 1015 1016：裸01背包+1 1 #include 2 #include 3 using namespace std; 4 int f[20001],a[31],n,v; 5 int main(){ 6 cin>>v>>n; 7 for (int i=1; i<=n; i++) scanf("%d",&a[i]); 8 for (int i=1; i<=n; i++) 9 for (int j=v; j>=a[i]; j--) 10 f[j]=max(f[j],a[i]+f[j-a[i]]); 11 cout< 12 } 1016 1017：并查集之后统计没有用的合并次数即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 int fa[101],n,m,ans; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 inline int find(int x){ 28 fa[x]=fa[x]==x?x:find(fa[x]); 29 return fa[x]; 30 } 31 int main(){ 32 n=in(); m=in(); 33 For(i,1,m) fa[i]=i; 34 For(i,1,n){ 35 int x=in(),y=in(); 36 if (find(x)==find(y)) ans++; 37 else fa[find(x)]=find(y); 38 } 39 printf("%d",ans); 40 } 1017 1018:20！用LL就可以存了。。。。然后按题意模拟一下即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 int n,k,mo=1,a[30]; 20 ll ans=1; 21 inline int in(){ 22 int x=0,f=1; 23 char ch=getchar(); 24 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 25 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 26 return x*f; 27 } 28 int main(){ 29 n=in();k=in(); 30 bool p=0; 31 For(i,1,k) mo*=10; 32 For(i,1,n) 33 ans*=i; 34 while(ans%10==0) ans/=10; 35 int cnt=0; 36 while(ans){ 37 a[++cnt]=ans%10; 38 ans/=10; 39 } 40 Ford(i,min(cnt,k),1) printf("%d",a[i]); 41 } 1018 1019：通过进行数学推导，我们容易发现，对于b[i] 中大于a[i]的数，用它们任意进行相减的和是一定的，反之亦然，于是我们只需要将b数组与a数组按大小一一匹配即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 int a[10001],b[10001],n,ans; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 int main(){ 28 n=in(); 29 For(i,1,n) a[i]=in(); 30 For(i,1,n) b[i]=in(); 31 sort(a+1,a+n+1); 32 sort(b+1,b+n+1,greater 33 For(i,1,n) 34 ans+=abs(b[i]-a[i]); 35 printf("%d",ans); 36 } 1019 1020：筛法筛质数，然后对于每个a[i]暴力枚举注意及时剪枝就行了。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 bool f[20001]; 20 int pr[20001],ma,maxx,n,x,cnt; 21 inline int in(){ 22 int x=0,f=1; 23 char ch=getchar(); 24 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 25 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 26 return x*f; 27 } 28 void init(){ 29 f[1]=1; 30 For(i,2,20000) 31 if (!f[i]) 32 For(j,2,20000/i) 33 f[i*j]=1; 34 For(i,2,20000) 35 if (!f[i]) 36 pr[++cnt]=i; 37 return; 38 } 39 int main(){ 40 init(); 41 n=in(); 42 For(i,1,n){ 43 x=in(); 44 Ford(i,cnt,1) 45 if (pr[i] 46 else if (x%pr[i]==0){ 47 ma=x; 48 maxx=pr[i]; 49 } 50 } 51 printf("%d",ma); 52 } 1020 1021:模拟即可。。。。 1021 1022：进制转换你不会吗？（从低位向上除，判断是否是奇数即可） 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 ll x; 20 bool ans[33]; 21 inline ll in(){ 22 ll x=0,f=1; 23 char ch=getchar(); 24 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 25 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 26 return x*f; 27 } 28 int main(){ 29 x=in(); 30 ll p=1; 31 int i=0; 32 for(p=1; i<=32; p*=(-2),i++) 33 if ((x/p)&1){ 34 ans[i]=1; 35 x-=p; 36 } 37 while(!ans[i]&&i) i--; 38 Ford(j,i,0) 39 printf("%d",(ans[j]?1:0)); 40 } 1022 1023:简单DP，用f[i][j]表示第i分钟时疲倦值为j的最大距离，则f[i][j]=f[i-1][j-1]+d[i](1<=j<=m),f[i][0]=max(f[i-1][0],f[i-j][j])(1<=j<=i&&j<=m) 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 int d[2001],n,m,f[2001][501]; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 int main(){ 28 n=in(),m=in(); 29 For(i,1,n) d[i]=in(); 30 For(i,1,n){ 31 f[i][0]=f[i-1][0]; 32 For(j,1,min(i,m)) f[i][0]=max(f[i][0],f[i-j][j]); 33 For(j,1,m) f[i][j]=f[i-1][j-1]+d[i]; 34 } 35 printf("%d",f[n][0]); 36 } 1023 1024:最长不下降子序列还是很简单的吧。。。注意一下输入的处理就行了。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 int dy[26],a[255],f[255],cnt,ans[255]; 20 char zxy[255]; 21 inline int in(){ 22 int x=0,f=1; 23 char ch=getchar(); 24 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 25 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 26 return x*f; 27 } 28 int doit(int n){ 29 mem(f,0); 30 f[1]=0; 31 int ans=0; 32 For(i,2,n) 33 For(j,1,i-1) 34 if (a[i]>=a[j]){ 35 f[i]=max(f[i],f[j]+1); 36 ans=max(f[i],ans); 37 } 38 return ans; 39 } 40 int main(){ 41 scanf("%s",zxy); 42 For(i,0,strlen(zxy)-1) 43 dy[zxy[i]-'a']=i; 44 while(scanf("%s",zxy+1)!=EOF){ 45 cnt=0; 46 For(i,0,strlen(zxy)-1) 47 a[++cnt]=dy[zxy[i]-'a']; 48 printf("%d",doit(cnt)); 49 } 50 } 1024 1025：判断奇偶性你不会吗？ 1 #include 2 #include 3 #include 4 using namespace std; 5 char a[255],b; 6 int n; 7 int main(){ 8 cin>>n; 9 getchar(); 10 for (int i=1; i<=n; i++){ 11 gets(a); 12 b=a[strlen(a)-1]; 13 if (b%2==0) cout<<"even\n"; 14 else cout<<"odd\n"; 15 } 16 } 1025 1026:数据范围这么小，打暴力不就好了。。。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 bool f[241][241]; 20 int n,m,I,ans; 21 inline int in(){ 22 int x=0,f=1; 23 char ch=getchar(); 24 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 25 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 26 return x*f; 27 } 28 int main(){ 29 n=in(),m=in(),I=in(); 30 int x,xx,y,yy; 31 For(i,1,I){ 32 x=in(),y=in(),xx=in(),yy=in(); 33 For(i,x,xx) 34 For(j,y,yy) 35 f[i][j]=1; 36 } 37 For(i,1,n) 38 For(j,1,m) 39 if (f[i][j]) ans++; 40 printf("%d",ans); 41 } 1026 1027：按照题意搜索即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 int a[45][45],dx[4]={1,0,-1,0},dy[4]={0,1,0,-1},n,m; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 int bfs(){ 28 int x=1,y=1; 29 int ans=a[1][1]; 30 a[1][1]=0; 31 while(x!=n||y!=m){ 32 int ma=0,nx,ny; 33 For(i,0,3) 34 if (a[x+dx[i]][y+dy[i]]>ma){ 35 ma=a[x+dx[i]][y+dy[i]]; 36 nx=x+dx[i],ny=y+dy[i]; 37 } 38 ans+=ma; 39 x=nx,y=ny; 40 a[x][y]=0; 41 } 42 return ans; 43 } 44 int main(){ 45 n=in(),m=in(); 46 For(i,1,n) 47 For(j,1,m) 48 a[i][j]=in(); 49 printf("%d",bfs()); 50 } 1027 1028：又一次的01背包。。 1 #include 2 #include 3 using namespace std; 4 int f[45001],a[501],n,v; 5 inline int in(){ 6 int x=0,f=1; 7 char ch=getchar(); 8 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 9 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 10 return x*f; 11 } 12 int main(){ 13 v=in(),n=in(); 14 for (int i=1; i<=n; i++) a[i]=in(); 15 for (int i=1; i<=n; i++) 16 for (int j=v; j>=a[i]; j--) 17 f[j]=max(f[j],a[i]+f[j-a[i]]); 18 printf("%d",f[v]); 19 } 1028 1029：暴力枚举答案检查就行了吧（你要无聊加个二分也行= =） 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 string a,b; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 bool check(string a,string b,int l){ 28 For(i,0,l-1) 29 if (a[i]!=b[b.size()-l+i]) return 0; 30 return 1; 31 } 32 int main(){ 33 cin>>a;cin>>b; 34 Ford(i,min(a.size(),b.size()),1) 35 if (check(a,b,i)||check(b,a,i)){ 36 printf("%d",i); 37 break; 38 } 39 } 1029 1030：BFS经典题 1 #include 2 #include 3 using namespace std; 4 struct fff{ 5 int nn,x,y; 6 }a[100001]; 7 int dx[9]={0,1,-1,1,-1,1,-1,0,0},dy[9]={0,0,0,1,1,-1,-1,1,-1},h,t,n,m,x,y; 8 char aa[101]; 9 bool b[101][101]; 10 int main(){ 11 cin>>n>>m>>y>>x; 12 for (int i=m; i>=1; i--){ 13 cin>>aa; 14 for (int j=0; j 15 b[i][j+1]=aa[j]=='.'?0:1; 16 } 17 h=0,t=1; 18 a[1].x=x; 19 a[1].y=y; 20 a[1].nn=0; 21 b[x][y]=1; 22 do{ 23 h++; 24 for (int i=1; i<=8; i++){ 25 x=a[h].x+dx[i]; 26 y=a[h].y+dy[i]; 27 if (x>0&&x<=m&&y>0&&y<=n&&!b[x][y]){ 28 t++; 29 a[t].nn=a[h].nn+1; 30 a[t].x=x; 31 a[t].y=y; 32 b[x][y]=1; 33 } 34 } 35 }while(h 36 cout< 37 } 1030 1031:典型最短路，我用的是SPFA：） 1 #include 2 #include 3 #include 4 using namespace std; 5 int dis[2501],n,m,s,e,x,y,z,a[2501][101][3],pp[100001],h,t; 6 bool b[3001]; 7 int main(){ 8 cin>>n>>m>>s>>e; 9 for (int i=1; i<=m; i++){ 10 scanf("%d%d%d",&x,&y,&z); 11 a[x][++a[x][0][0]][1]=z; 12 a[x][a[x][0][0]][2]=y; 13 a[y][++a[y][0][0]][1]=z; 14 a[y][a[y][0][0]][2]=x; 15 } 16 h=0,t=1; 17 pp[1]=s; 18 b[s]=1; 19 20 memset(dis,127/3,sizeof(dis)); 21 dis[s]=0; 22 do{ 23 h++; 24 for (int i=1; i<=a[pp[h]][0][0]; i++) 25 if (dis[a[pp[h]][i][2]]>dis[pp[h]]+a[pp[h]][i][1]){ 26 dis[a[pp[h]][i][2]]=dis[pp[h]]+a[pp[h]][i][1]; 27 if (!b[a[pp[h]][i][2]]){ 28 t++; 29 pp[t]=a[pp[h]][i][2]; 30 b[a[pp[h]][i][2]]=1; 31 } 32 } 33 b[pp[h]]=0; 34 }while(h 35 cout< 36 } 1031 1032:由于大面额的钱币是小面额的倍数，我们容易想到一个贪心思路：先拿大面额凑，再拿小面额凑，最后注意一下钱不够和最大面额大于所需支付工资的情况即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 struct zxy{ 20 int n,v; 21 }a[21]; 22 int n,v,ans; 23 inline int in(){ 24 int x=0,f=1; 25 char ch=getchar(); 26 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 27 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 28 return x*f; 29 } 30 bool cmp(zxy a,zxy b){ 31 return a.v 32 } 33 int main(){ 34 n=in(),v=in(); 35 For(i,1,n) 36 a[i].v=in(),a[i].n=in(); 37 sort(a+1,a+n+1,cmp); 38 Ford(i,n,1) 39 if(a[i].v>v){ 40 n--; 41 ans+=a[i].n; 42 } 43 while(1){ 44 int nt=0; 45 Ford(i,n,1) 46 while(a[i].v+nt 47 nt+=a[i].v,a[i].n--; 48 For(i,1,n) 49 if(a[i].n){ 50 nt+=a[i].v; 51 a[i].n--; 52 break; 53 } 54 if (nt 55 ans++; 56 } 57 printf("%d",ans); 58 } 1032 1033：题意就是叫你求一颗二叉树的深度，DP或者dfs一下就行了，时间效率O（n）。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 using namespace std; 19 struct zxy{ 20 int lc,rc,deep; 21 }a[1001]; 22 int n; 23 int dfs(int k){ 24 a[a[k].lc].deep=a[a[k].rc].deep=a[k].deep+1; 25 if (!a[k].lc&&!a[k].rc) return a[k].deep; 26 if (!a[k].rc) return dfs(a[k].lc); 27 if (!a[k].lc) return dfs(a[k].rc); 28 return max(dfs(a[k].lc),dfs(a[k].rc)); 29 } 30 inline int in(){ 31 int x=0,f=1; 32 char ch=getchar(); 33 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 34 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 35 return x*f; 36 } 37 int main(){ 38 n=in(); 39 int x,y,z; 40 For(i,1,n-1){ 41 x=in(),y=in(),z=in(); 42 a[x].lc=y,a[x].rc=z; 43 } 44 a[1].deep=1; 45 printf("%d",dfs(1)); 46 } 1033 1034：按题意DP或者打个最短路就行了（最短路待更新），DP按题意做就好 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define max(a,b) (a 15 #define min(a,b) (a 16 #define ll long long 17 #define mod 1000000007 18 #define INF 2000000000 19 using namespace std; 20 int f[10002],c[10001],s[10001],n,k; 21 inline int in(){ 22 int x=0,f=1; 23 char ch=getchar(); 24 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 25 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 26 return x*f; 27 } 28 int main(){ 29 n=in(),k=in(); 30 For(i,1,k) s[i]=in(),c[i]=in(); 31 For(i,2,n) f[i]=-INF; 32 For(i,1,n){ 33 bool b=0; 34 For(j,1,k) 35 if (s[j]==i){ 36 f[i+c[j]]=max(f[i+c[j]],f[i]); 37 b=1; 38 } 39 if (!b) f[i+1]=max(f[i+1],f[i]+1); 40 } 41 printf("%d",f[n+1]); 42 } 1034DP 1035：原谅本人过蒻并不会二分图匹配。。。（容我以后填坑） 1036：排序以后处理一下即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 int n,a[200001],b[200001]; 19 inline int in(){ 20 int x=0,f=1; 21 char ch=getchar(); 22 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 23 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 24 return x*f; 25 } 26 int main(){ 27 n=in(); 28 For(i,1,n) a[i]=in(); 29 sort(a+1,a+n+1); 30 int cnt=1; 31 For(i,2,n) 32 if (a[i]!=a[i-1]){ 33 printf("%d %d\n",a[i-1],cnt); 34 cnt=1; 35 } 36 else ++cnt; 37 printf("%d %d",a[n],cnt); 38 } 1036 1037：高精乘法维护抹零后的后20位即可，其余同1018 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 struct zxy{ 19 int begin,l,v[500001]; 20 zxy(){ 21 begin=1,l=1; 22 v[1]=1; 23 } 24 void friend operator *(zxy &a,int b){ 25 For(i,a.begin,min(a.begin+20,a.l)) 26 a.v[i]*=b; 27 For(i,a.begin,min(a.begin+20,a.l)){ 28 a.v[i+1]+=a.v[i]/10; 29 a.v[i]%=10; 30 if(a.v[a.l+1])a.l++; 31 while(!a.v[a.begin])a.begin++; 32 } 33 } 34 }ans; 35 int n,k; 36 inline int in(){ 37 int x=0,f=1; 38 char ch=getchar(); 39 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 40 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 41 return x*f; 42 } 43 int main(){ 44 n=in(),k=in(); 45 For(i,1,n) ans*i; 46 Ford(i,ans.begin+k-1,ans.begin) 47 printf("%d",ans.v[i]); 48 } 1037 1038：线段树裸题（大神可以用RMQ，本人太蒻） 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 #define mid ((l+r)>>1) 18 using namespace std; 19 struct zxy{ 20 int mi; 21 }tr[400001]; 22 int n,q; 23 inline int in(){ 24 int x=0,f=1; 25 char ch=getchar(); 26 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 27 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 28 return x*f; 29 } 30 inline void update(int l,int r,int k,int t,int add){ 31 if (l==r) { 32 tr[k].mi=add; 33 return; 34 } 35 if (t<=mid) update(l,mid,k<<1,t,add); 36 else update(mid+1,r,k<<1|1,t,add); 37 tr[k].mi=min(tr[k<<1].mi,tr[k<<1|1].mi); 38 return; 39 } 40 inline int query(int l,int r,int a,int b,int k){ 41 if (a==l&&b==r) return tr[k].mi; 42 if (b<=mid) return query(l,mid,a,b,k<<1); 43 if (a>mid) return query(mid+1,r,a,b,k<<1|1); 44 return min(query(l,mid,a,mid,k<<1),query(mid+1,r,mid+1,b,k<<1|1)); 45 } 46 int main(){ 47 n=in(),q=in(); 48 For(i,1,n) update(1,n,1,i,in()); 49 For(i,1,q){ 50 int x=in(),y=in(); 51 printf("%d ",query(1,n,x,y,1)); 52 } 53 } 1038 1039：线段树裸题+1 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 #define mid ((l+r)>>1) 18 using namespace std; 19 struct zxy{ 20 int mi; 21 }tr[400001]; 22 int n,q; 23 inline int in(){ 24 int x=0,f=1; 25 char ch=getchar(); 26 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 27 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 28 return x*f; 29 } 30 inline void update(int l,int r,int k,int t,int add){ 31 if (l==r) { 32 tr[k].mi=add; 33 return; 34 } 35 if (t<=mid) update(l,mid,k<<1,t,add); 36 else update(mid+1,r,k<<1|1,t,add); 37 tr[k].mi=min(tr[k<<1].mi,tr[k<<1|1].mi); 38 return; 39 } 40 inline int query(int l,int r,int a,int b,int k){ 41 if (a==l&&b==r) return tr[k].mi; 42 if (b<=mid) return query(l,mid,a,b,k<<1); 43 if (a>mid) return query(mid+1,r,a,b,k<<1|1); 44 return min(query(l,mid,a,mid,k<<1),query(mid+1,r,mid+1,b,k<<1|1)); 45 } 46 int main(){ 47 n=in(),q=in(); 48 For(i,1,n) update(1,n,1,i,in()); 49 For(i,1,q){ 50 int k=in(),x=in(),y=in(); 51 if (!(k^1)) 52 printf("%d ",query(1,n,x,y,1)); 53 else update(1,n,1,x,y); 54 } 55 } 1039 1040&1041：裸的高精加减法，注意一下字符串处理的一些细节即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 struct data{ 19 int l,v[2001]; 20 void clear(){ 21 l=0; 22 mem(v,0); 23 } 24 data friend operator +(data a,data b){ 25 data c;c.clear(); 26 c.l=max(a.l,b.l); 27 For(i,1,c.l) c.v[i]=a.v[i]+b.v[i]; 28 For(i,1,c.l) c.v[i+1]+=c.v[i]/10,c.v[i]%=10; 29 if (c.v[c.l+1]) ++c.l; 30 return c; 31 } 32 data friend operator -(data a,data b){ 33 data c;c.clear(); 34 c.l=a.l; 35 For(i,1,c.l) 36 if (a.v[i]>=b.v[i]) 37 c.v[i]=a.v[i]-b.v[i]; 38 else { 39 a.v[i+1]--; 40 c.v[i]=a.v[i]+10-b.v[i]; 41 } 42 while(!c.v[c.l]&&c.l) c.l--; 43 return c; 44 } 45 }ans,tmp[260]; 46 char a[260],ch[260]; 47 int cnt; 48 inline int in(){ 49 int x=0,f=1; 50 char ch=getchar(); 51 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 52 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 53 return x*f; 54 } 55 int main(){ 56 scanf("%s",a); 57 cnt=1; 58 Ford(i,strlen(a)-1,0) 59 if (a[i]!='+'&&a[i]!='-') tmp[cnt].v[++tmp[cnt].l]=a[i]-'0'; 60 else ch[cnt++]=a[i]; 61 ans=tmp[cnt]; 62 Ford(i,cnt-1,1) 63 if (ch[i]=='+') ans=ans+tmp[i]; else ans=ans-tmp[i]; 64 Ford(i,ans.l,1) printf("%d",ans.v[i]); 65 } 1041&1042 1042&1043：栈的练习。。。。注意处理细节。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #include 11 #define For(i,a,b) for (int i=a; i<=b; i++) 12 #define Ford(i,a,b) for (int i=a; i>=b; i--) 13 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 14 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 15 #define ll long long 16 #define mod 1000000007 17 #define INF 2000000000 18 using namespace std; 19 stack 20 stack 21 string s; 22 int cnt; 23 inline int in(){ 24 int x=0,f=1; 25 char ch=getchar(); 26 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 27 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 28 return x*f; 29 } 30 int inline getlevel(char x){ 31 if (x=='+'||x=='-') return 1; 32 if (x=='*'||x=='/') return 2; 33 if (x=='^') return 3; 34 if (x=='(') return 100; 35 if (x=='#') return -INF; 36 return -100; 37 } 38 ll calc(){ 39 char tmp=ch.top();ch.pop(); 40 ll b=num.top();num.pop(); 41 ll a=num.top();num.pop(); 42 if(tmp=='+')return a+b; 43 if(tmp=='-')return a-b; 44 if(tmp=='*')return a*b; 45 if(tmp=='/')return a/b; 46 ll ans=1; 47 for(int i=1;i<=b;i++) 48 ans*=a; 49 return ans; 50 } 51 int main(){ 52 cin>>s; 53 For(i,0,s.size()-1) 54 if (s[i]=='(') cnt++; 55 else if (s[i]==')') --cnt; 56 while(cnt>0) cnt--,s=s+")"; 57 while(cnt<0) cnt++,s="("+s; 58 s=s+"#"; 59 ch.push('#'); 60 num.push(0); 61 int i=0; 62 while(s[i]!='#'||ch.top()!='#'){ 63 if (s[i]<'0'||s[i]>'9'){ 64 if(getlevel(s[i])>getlevel(ch.top())){ 65 s[i]=s[i]=='('?'[':s[i]; 66 ch.push(s[i]);i++; 67 } 68 else 69 if (ch.top()=='[') ch.pop(),i++; 70 else num.push(calc()); 71 } 72 else{ 73 ll x=0; 74 while(s[i]>='0'&&s[i]<='9') x=x*10+s[i++]-'0'; 75 num.push(x); 76 } 77 } 78 cout< 79 } 1042&1043 1044：递推 #include #include #define For(i,a,b) for (int i=1; i<=a; i+=b) #define in(a) scanf("%d",&a) using namespace std; int n,f[26][26],a[26][26],ans=-0x7ffffff; int main(){ in(n); For(i,n,1) For(j,i,1) in(a[i][j]); f[1][1]=a[1][1]; for (int i=2; i<=n; i++) For(j,i,1){ f[i][j]=max(f[i-1][j],f[i-1][j-1])+a[i][j]; } For(i,n,1) ans=max(ans,f[n][i]); cout< } 1044 1045：区间DP，f[i][j][k]表示在[i,j]内用了k个乘号,f[l][r][k]=max(f[l][i][j]+f[i+1][r][k-j],f[l][i][j]*f[i+1][r][k-j-1])(1<=l,r<=n;l<=i 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 int f[101][101][101],a[101],n,num; 19 inline int in(){ 20 int x=0,f=1; 21 char ch=getchar(); 22 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 23 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 24 return x*f; 25 } 26 inline int dp(int l,int r,int k){ 27 if (r 28 if (f[l][r][k]>0) return f[l][r][k]; 29 if (r-l==k){ 30 f[l][r][k]=1; 31 For(i,l,r) 32 f[l][r][k]*=a[i]; 33 return f[l][r][k]; 34 } 35 For(i,l,r-1) 36 For(j,0,k) 37 f[l][r][k]=max(f[l][r][k],max(dp(l,i,j)+dp(i+1,r,k-j),dp(l,i,j)*dp(i+1,r,k-j-1))); 38 return f[l][r][k]; 39 } 40 int main(){ 41 n=in(),num=in(); 42 mem(f,(-1)); 43 For(i,1,n) f[i][i][0]=a[i]=in(); 44 printf("%d",dp(1,n,num)); 45 } 1045 1046:DP,用f[i][j]表示a串排了i个，b串拍了j个，状态转移方程是f[i][j]=min(f[i-1][j-1]+abs(a[i-1]-b[j-1]),min(f[i-1][j]+k,f[i][j-1]+k))，边界条件是f[i][0]=i*k,f[0][j]=j*k。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 #define MAXN 2001 18 using namespace std; 19 int f[MAXN][MAXN],n,m,k; 20 char a[MAXN],b[MAXN]; 21 inline int in(){ 22 int x=0,f=1; 23 char ch=getchar(); 24 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 25 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 26 return x*f; 27 } 28 int dp(){ 29 mem(f,127/3); 30 f[0][0]=0; 31 For(i,1,n) f[i][0]=i*k; 32 For(i,1,m) f[0][i]=i*k; 33 For(i,1,n) 34 For(j,1,m) 35 f[i][j]=min(f[i-1][j-1]+abs(a[i-1]-b[j-1]),min(f[i-1][j]+k,f[i][j-1]+k)); 36 return f[n][m]; 37 } 38 int main(){ 39 scanf("%s%s%d",a,b,&k); 40 n=strlen(a); 41 m=strlen(b); 42 printf("%d",dp()); 43 } 1046 1048：我们不妨假设齐王的马是从快到小依次出的，用f[l][r]表示田忌可以从自己第l强到第r蒻的马中任选一匹在本轮出战，此时赛马已经比过了(n-r+l-1)轮,我们容易想到一个DP的状态方程式： f[l][r]=max(f[l+1][r]+price(l,n-r+l),f[l][r-1]+price(r,n-r+l));price(i,j)表示用田忌的第i匹马和第j匹马在本轮出战能拿到的money。然后用记忆化搜索实现即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 1000000000 17 using namespace std; 18 int f[1001][1001],n,a[1001],b[1001]; 19 inline int in(){ 20 int x=0,f=1; 21 char ch=getchar(); 22 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 23 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 24 return x*f; 25 } 26 inline int price(int x,int y){ 27 if (a[x]>b[y]) return 200; 28 if (a[x] 29 return 0; 30 } 31 inline int dp(int l,int r){ 32 if (f[l][r]>(-INF)) return f[l][r]; 33 if (r 34 if (l==r) { 35 f[l][r]=price(l,n); 36 return f[l][r]; 37 } 38 f[l][r]=max(dp(l+1,r)+price(l,n-r+l),dp(l,r-1)+price(r,n-r+l)); 39 return f[l][r]; 40 } 41 int main(){ 42 mem(f,-127/2); 43 n=in(); 44 For(i,1,n) a[i]=in(); 45 For(i,1,n) b[i]=in(); 46 sort(a+1,a+n+1,greater 47 sort(b+1,b+n+1,greater 48 printf("%d",dp(1,n)); 49 } 1048 1049：裸DP题，方程式都没什么好讲的。。。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 int f[5001],a[5001],n,ans; 19 inline int in(){ 20 int x=0,f=1; 21 char ch=getchar(); 22 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 23 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 24 return x*f; 25 } 26 int main(){ 27 n=in(); 28 For(i,1,n) a[i]=in(); 29 f[0]=0; 30 For(i,1,n) 31 For(j,0,i-1) 32 if (a[i]>=a[j]){ 33 f[i]=max(f[i],f[j]+1); 34 ans=max(f[i],ans); 35 } 36 printf("%d",ans); 37 } 1049 1050：用f[i][j]表示a串前i个字符与b串前j个字符进行匹配的最长子串长度，状态转移方程式是：f[i][j]=max(f[i-1][j],f[i][j-1])(a[i]!=b[j])&f[i][j]=f[i-1][j-1]+1(a[i]=a[j]).然后用记忆化搜索实现即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 int f[2001][2001]; 19 string a,b; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 inline int dp(int i,int j){ 28 if (i<0||j<0) return 0; 29 if (f[i][j]) return f[i][j]; 30 if (a[i]==b[j]) return f[i][j]=dp(i-1,j-1)+1; 31 return f[i][j]=max(dp(i-1,j),dp(i,j-1)); 32 } 33 int main(){ 34 cin>>a>>b; 35 printf("%d",dp(a.size()-1,b.size()-1)); 36 } 1050 1051:典型的树形DP题，我们采用左儿子右兄弟法存树，f[i][j]表示第i个节点还能选j门课的最优值，枚举DP到儿子时可以选的课数l(1<=l<=j)，可以发现 f[i][j]=max(f[儿子][j-i-1]+f[兄弟][i]+i节点的权值,f[兄弟][j]);然后我们使用记忆化搜索实现树上DP即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 struct zxy{ 19 int v,son,bra; 20 }tr[301]; 21 int n,f[301][301],m,ans; 22 inline int in(){ 23 int x=0,f=1; 24 char ch=getchar(); 25 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 26 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 27 return x*f; 28 } 29 inline int dp(int k,int t){ 30 if (!(k&&t)) return 0; 31 if (f[k][t]) return f[k][t]; 32 f[k][t]=tr[k].v; 33 For(i,1,t) 34 f[k][t]=max(f[k][t],dp(tr[k].son,t-i)+dp(tr[k].bra,i-1)+tr[k].v); 35 f[k][t]=max(dp(tr[k].bra,t),f[k][t]); 36 return f[k][t]; 37 } 38 int main(){ 39 n=in(),m=in(); 40 For(i,1,n){ 41 int x=in(),y=in(); 42 tr[i].bra=tr[x].son; 43 tr[x].son=i; 44 tr[i].v=y; 45 } 46 f[0][0]=0; 47 tr[0].v=0; 48 printf("%d",dp(tr[0].son,m)); 49 } 1051 1052：显然又是一题树上DP，用f[i][0/1]表示取不取第i节点时的最优状态，采用了很暴力的方法存树，找到root，跑一次dfs枚举儿子累加状态即可。 状态转移方程：f[i][0]=Σmax(f[儿子][1],f[儿子][0])，f[i][1]=第i点的权值+∑f[儿子][0]. 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 int n,f[6001][2],fa[6001]; 19 bool b[6001]; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 inline void treedp(int k){ 28 b[k]=0; 29 For(i,1,n) 30 if (b[i]&&fa[i]==k){ 31 treedp(i); 32 f[k][1]+=f[i][0]; 33 f[k][0]+=max(f[i][1],f[i][0]); 34 } 35 } 36 int main(){ 37 n=in(); 38 For(i,1,n) f[i][1]=in(); 39 For(i,1,n-1){ 40 int x=in(),y=in(); 41 fa[x]=y; 42 b[x]=1; 43 } 44 int s; 45 For(i,1,n) if (!b[i]){ 46 s=i; 47 break; 48 } 49 treedp(s); 50 printf("%d",max(f[s][0],f[s][1])); 51 } 1052 1053：字符串处理，按题意做，注意连续'-'和开头以及结尾的处理就行了。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (char i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i<=b; i++) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 string a; 19 int p1,p2,p3; 20 inline int in(){ 21 int x=0,f=1; 22 char ch=getchar(); 23 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 24 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 25 return x*f; 26 } 27 inline bool digit(char k){ 28 return k>='0'&&k<='9'; 29 } 30 inline bool alpha(char k){ 31 return k>='a'&&k<='z'; 32 } 33 inline void get_(char a,char b){ 34 if (b<=a||(alpha(a)&&digit(b))||(digit(a)&&alpha(b))||a=='-'||b=='-') { 35 putchar('-'); 36 return; 37 } 38 int cnt=0; 39 if (p3-1){ 40 if (p1==3) For(i,a+1,b-1) 41 Ford(j,1,p2) 42 putchar('*'); 43 if (p1==2&&alpha(a)) 44 for(char i=b-1; i>a; i--) 45 Ford(j,1,p2) 46 putchar(i-'a'+'A'); 47 if (p1==1||(p1==2&&digit(a))) for(char i=b-1; i>a; i--) 48 Ford(j,1,p2) 49 putchar(i); 50 return; 51 } 52 if (p1==3) For(i,a+1,b-1) 53 Ford(j,1,p2) 54 putchar('*'); 55 if (p1==2&&alpha(a)) 56 For(i,a+1,b-1) 57 Ford(j,1,p2) 58 putchar(i-'a'+'A'); 59 if (p1==1) For(i,a+1,b-1) 60 Ford(j,1,p2) 61 putchar(i); 62 } 63 int main(){ 64 p1=in(),p2=in(),p3=in(); 65 cin>>a; 66 Ford(i,0,a.size()-1) 67 if (a[i]!='-'||i==0) putchar(a[i]); 68 else get_(a[i-1],a[i+1]); 69 } 1053 1055:又是区间DP,用f[l][r]表示将[l,r]合并成一堆以后的最优值，这样的话状态转移方程式是：f[l][r]=f[l][i]+f[i+1][r]+∑(j=l,j<=r)a[j](l<=i<=r)，其中我们可以用前缀和简化∑，然后使用记忆化搜索即可。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 100000000 17 using namespace std; 18 int f[301][301],n,s[301]; 19 inline int in(){ 20 int x=0,f=1; 21 char ch=getchar(); 22 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 23 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 24 return x*f; 25 } 26 inline int dp(int l,int r){ 27 if (f[l][r] 28 if (l==r) return 0; 29 For(i,l,r-1) f[l][r]=min(dp(l,i)+dp(i+1,r)+s[r]-s[l-1],f[l][r]); 30 return f[l][r]; 31 } 32 int main(){ 33 n=in(); 34 For(i,1,n) s[i]=s[i-1]+in(); 35 mem(f,127/3) 36 printf("%d",dp(1,n)); 37 } 1055 1056：still区间DP，首先对循环这个特性我们进行处理，用[n+1,2*n]复制与[1,n]的相同信息，这样之后从1~n都进行一次长度为n的DP取最优即可。 DP的方法：用f[l][r]表示将[l,r]合并后的最优值，状态转移方程：f[l][r]=f[l][i]+f[i+1][r]+a[l]*a[i+1]*a[r](l<=i<=r)，边界是f[i][i]=0； 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 int f[205][205],a[205],n,ans; 19 inline int in(){ 20 int x=0,f=1; 21 char ch=getchar(); 22 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 23 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 24 return x*f; 25 } 26 inline int dp(int l,int r){ 27 if (f[l][r]) return f[l][r]; 28 if (l==r) return 0; 29 For(i,l,r-1) f[l][r]=max(f[l][r],dp(l,i)+dp(i+1,r)+a[l]*a[i+1]*a[r+1]); 30 return f[l][r]; 31 } 32 int main(){ 33 n=in(); 34 For(i,1,n) a[i]=in(); 35 For(i,n+1,2*n) a[i]=a[i-n]; 36 For(i,1,n) 37 ans=max(ans,dp(i,i+n-1)); 38 printf("%d",ans); 39 } 1056 1057：裸01背包，处理一下依附关系即可，具体参见程序。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 struct zxy{ 19 int ls,rs,v,imp; 20 bool pp; 21 }tr[61]; 22 int n,m,f[3201]; 23 inline int in(){ 24 int x=0,f=1; 25 char ch=getchar(); 26 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 27 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 28 return x*f; 29 } 30 int main(){ 31 m=in(),n=in(); 32 m/=10; 33 For(i,1,n){ 34 tr[i].v=in(); 35 tr[i].v/=10; 36 tr[i].imp=in(); 37 int y=in(); 38 if (y){ 39 tr[i].pp=1; 40 if (tr[y].ls) tr[y].rs=i; 41 else tr[y].ls=i; 42 } 43 } 44 tr[0].v=tr[0].imp=0; 45 For(i,1,n) 46 if (!tr[i].pp) 47 Ford(j,m,tr[i].v){ 48 int ls=tr[i].ls,rs=tr[i].rs; 49 if (j-tr[i].v>=tr[ls].v) 50 f[j]=max(f[j],f[j-tr[i].v-tr[ls].v]+tr[i].v*tr[i].imp+tr[ls].v*tr[ls].imp); 51 if (j-tr[i].v>=tr[rs].v) 52 f[j]=max(f[j],f[j-tr[i].v-tr[rs].v]+tr[i].v*tr[i].imp+tr[rs].v*tr[rs].imp); 53 if (j-tr[i].v>=tr[rs].v+tr[ls].v) 54 f[j]=max(f[j],f[j-tr[i].v-tr[rs].v-tr[ls].v]+tr[i].v*tr[i].imp+tr[rs].v*tr[rs].imp+tr[ls].v*tr[ls].imp); 55 f[j]=max(f[j],f[j-tr[i].v]+tr[i].v*tr[i].imp); 56 } 57 printf("%d",f[m]*10); 58 } 1057 1058：单纯的模拟，题意自己去外面找去┑(￣Д ￣)┍ 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 struct zxy{ 19 int no,no2; 20 }order[401]; 21 int use[21][21],n,m,cnt[21],t[21][21],mac[21][21],mact[21],ans,end[21]; 22 bool used[21][401]; 23 inline int in(){ 24 int x=0,f=1; 25 char ch=getchar(); 26 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 27 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 28 return x*f; 29 } 30 inline bool pd(int ma,int s,int e){ 31 For(i,s,e) 32 if (used[ma][i]) return 0; 33 return 1; 34 } 35 int main(){ 36 m=in(),n=in(); 37 For(i,1,m*n) order[i].no=in(),order[i].no2=++cnt[order[i].no]; 38 For(i,1,n) 39 For(j,1,m) mac[i][j]=in(); 40 For(i,1,n) 41 For(j,1,m) t[i][j]=in(); 42 For(i,1,m*n){ 43 int gj=order[i].no,no=order[i].no2; 44 int tim=t[gj][no],macc=mac[gj][no]; 45 For(j,end[gj],INF) 46 if (pd(macc,j,j+tim-1)){ 47 For(k,j,j+tim-1) 48 used[macc][k]=1; 49 end[gj]=j+tim; 50 if (j+tim>ans) ans=j+tim; 51 break; 52 } 53 } 54 printf("%d",ans); 55 } 1058 1062:题目都说了和合并方法和合并傻子沙子差不多，这里就不多说怎么做了，做法与1055类似，详见代码 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 100000000 17 using namespace std; 18 int f[101][101],n,s[101],m; 19 inline int in(){ 20 int x=0,f=1; 21 char ch=getchar(); 22 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 23 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 24 return x*f; 25 } 26 inline int dpmi(int l,int r){ 27 if (f[l][r] 28 if (l==r) return 0; 29 For(i,l,r-1) f[l][r]=min(dpmi(l,i)+dpmi(i+1,r)+s[r]-s[l-1],f[l][r]); 30 return f[l][r]; 31 } 32 inline int dpma(int l,int r){ 33 if (f[l][r]) return f[l][r]; 34 if (l==r) return 0; 35 For(i,l,r-1) f[l][r]=max(dpma(l,i)+dpma(i+1,r)+s[r]-s[l-1],f[l][r]); 36 return f[l][r]; 37 } 38 int main(){ 39 n=in();m=in(); 40 For(i,1,n) s[i]=s[i-1]+in(); 41 int ma=dpma(1,n); 42 mem(f,127/3); 43 int mi=dpmi(1,n); 44 if (m 45 else if (m>ma) printf("It is easy"); 46 else printf("I will go to play WarIII"); 47 } 1062 1063：我们可以比较容易的想到一种贪心的方法，用头指针h和尾指针t来在数字串上模拟一个队列，对于队列判断一下是否符合条件，进行答案更新，这样的话效率还是很慢（O（mn）），我们可以考虑对队列中不同元素的数量进行一个统计，这样当cnt=m时，自然队列就符合条件，具体实现方法见代码，时间效率为O（n）。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 2000000000 17 using namespace std; 18 int n,m,a[200001],b[200001],h,t; 19 inline int in(){ 20 int x=0,f=1; 21 char ch=getchar(); 22 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 23 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 24 return x*f; 25 } 26 int main(){ 27 n=in(),m=in(); 28 For(i,1,n) a[i]=in(); 29 int cnt=0,ans=n+1; 30 h=1,t=0; 31 while(h<=n-m+1&&t<=n){ 32 while(cnt!=m&&t<=n) 33 cnt=b[a[++t]]?cnt:cnt+1,b[a[t]]++; 34 if (t-h+1 35 b[a[h]]--; 36 if (!b[a[h]]) cnt--; 37 h++; 38 } 39 if (ans<=n)printf("%d",ans); 40 else printf("NO"); 41 } 1063 1065：纯模拟 1 #include 2 #include 3 #include 4 using namespace std; 5 int n,c,x; 6 int main(){ 7 for (int i=1; i<=12; i++){ 8 n+=300; 9 scanf("%d",&x); 10 n-=x; 11 if (n<0){ 12 cout<<-i; 13 exit(0); 14 } 15 c+=n/100*120; 16 n%=100; 17 } 18 cout< 19 } 1065 1066：STL练习题，可以采用系统优先队列，自带堆等（本人采用自带堆排） 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #include 11 #define For(i,a,b) for (int i=a; i<=b; i++) 12 #define Ford(i,a,b) for (int i=a; i>=b; i--) 13 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 14 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 15 #define ll long long 16 #define mod 1000000007 17 #define INF 2000000000 18 using namespace std; 19 int heap[25001],hp,n,ans; 20 void in(int x){ 21 heap[++hp]=x; 22 push_heap(heap+1,heap+hp+1,greater 23 } 24 int out(){ 25 pop_heap(heap+1,heap+hp+1,greater 26 return heap[hp--]; 27 } 28 int main(){ 29 cin>>n; 30 for (int i=1; i<=n; i++){ 31 int x; 32 scanf("%d",&x); 33 in(x); 34 } 35 while(hp!=1){ 36 int x=out(),y=out(); 37 ans+=(x+y); 38 in(x+y); 39 } 40 cout< 41 } 1066 1067:用DP从前往后和从后往前各求一次到每个人的最长连续上升子序列长度，这样以来选每个人作为最高点时的答案就很好求了。 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #define For(i,a,b) for (int i=a; i<=b; i++) 11 #define Ford(i,a,b) for (int i=a; i>=b; i--) 12 #define File(fn) freopen(fn".in","r",stdin); freopen(fn".out","w",stdout); 13 #define mem(qaq,num) memset(qaq,num,sizeof(qaq)); 14 #define ll long long 15 #define mod 1000000007 16 #define INF 700000000 17 using namespace std; 18 int upp[102],n,a[102],dow[102],ans=0x7fffffff; 19 inline int in(){ 20 int x=0,f=1; 21 char ch=getchar(); 22 while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar(); 23 while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 24 return x*f; 25 } 26 int main(){ 27 n=in(); 28 For(i,1,n) a[i]=in(); 29 For(i,1,n) 30 For(j,0,i-1) 31 if (a[i]>a[j]) 32 upp[i]=max(upp[i],upp[j]+1); 33 Ford(i,n,1) 34 Ford(j,n+1,i+1) 35 if (a[i]>a[j]) 36 dow[i]=max(dow[i],dow[j]+1); 37 For(i,1,n) 38 ans=min(ans,n-upp[i]-dow[i]+1); 39 printf("%d",ans); 40 } 1067 1075：数字三角形无脑递推系列... 1 #include 2 #include 3 #define For(i,a,b) for (int i=1; i<=a; i+=b) 4 #define in(a) scanf("%d",&a) 5 using namespace std; 6 int n,a[26][26],ans=-32767;bool f[26][26][100]; 7 int main(){ 8 in(n); 9 For(i,n,1) 10 For(j,i,1) 11 in(a[i][j]); 12 f[1][1][a[1][1]%100]=1; 13 for (int i=2; i<=n; i++) 14 For(j,i,1){ 15 for (int k=0; k<=99; k++) if (f[i-1][j][k]) f[i][j][(k+a[i][j])%100]=1; 16 for (int k=0; k<=99; k++) if (f[i-1][j-1][k]) f[i][j][(k+a[i][j])%100]=1; 17 } 18 For(j,n,1) 19 for (int i=99; i>=0; i--) if (f[n][j][i]) ans=max(ans,i); 20 cout< 21 } 1075 1079：数字三角形无脑递推系列... 1 #include 2 #include 3 #define For(i,a,b) for (int i=a; i<=b; i++) 4 #define in(a) scanf("%d",&a) 5 using namespace std; 6 int n,f[26][26],a[26][26],ans=-0x7ffffff; 7 int main(){ 8 in(n); 9 For(i,1,n) 10 For(j,1,i) 11 in(a[i][j]); 12 f[1][1]=a[1][1]; 13 For(i,2,n/2) 14 For(j,1,i) 15 f[i][j]=max(f[i-1][j],f[i-1][j-1])+a[i][j]; 16 For(i,1,n/2-1) f[n/2][i]=0; 17 For(i,n/2+1,n) 18 For(j,n/2,i) 19 f[i][j]=max(f[i-1][j],f[i-1][j-1])+a[i][j]; 20 For(i,n/2,n) ans=max(ans,f[n][i]); 21 cout< 22 } 1079 1084：数字三角形无脑递推系列... 1 #include 2 #include 3 #define For(i,a,b) for (int i=a; i<=b; i++) 4 #define in(a) scanf("%d",&a) 5 using namespace std; 6 int n,f[26][26],a[26][26],ans=-0x7ffffff,x,y; 7 int main(){ 8 in(n); 9 For(i,1,n) 10 For(j,1,i) 11 in(a[i][j]); 12 in(x);in(y); 13 f[1][1]=a[1][1]; 14 For(i,2,x) 15 For(j,1,i) 16 f[i][j]=max(f[i-1][j],f[i-1][j-1])+a[i][j]; 17 For(i,1,x)if (i!=y) f[x][i]=-2147483647; 18 For(i,x+1,n) 19 For(j,y,i) 20 f[i][j]=max(f[i-1][j],f[i-1][j-1])+a[i][j]; 21 For(i,y,n) ans=max(ans,f[n][i]); 22 cout< 23 } 1084 本文由Melacau编写，Melacau代表M星向您问好，如果您不是在我的博客http://www.cnblogs.com/Melacau上看到本文，请您向我联系，email:13960948839@163.com.