#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
using namespace std;
int const N = 110;
int main()
{int s[N];int MAX = -1;int sum;int sum1;int n;int i,j,k;while(scanf("%d",&n)!=EOF){sum = 0;for(i=0; i<n; i++){scanf("%d",&s[i]);sum += s[i];}for(i=0; i<n; i++){for(j=i; j<n; j++){sum1 = sum;for(k=i; k<=j; k++){sum1 -= s[k];sum1 += (1-s[k]);if(sum1>MAX) MAX = sum1;}}}printf("%d\n",MAX);}return 0;
}

B. Hungry Sequence

special judge...

#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
using namespace std;int main()
{int n,i;cin >> n;for(i=n; i< 2*n ; ++i)printf("%d ", i);printf("\n");//printf("%d\n", 2*n-1);return 0;
}

C. Magic Five

这道题应该用快速幂来求，若是对于项数很多的等比数列，因为求和公式中包含了除号，所以不能直接取mod，应该进行快速幂的转化

例如求sum=2^1+2^2+2^3+2^4+2^5+2^6+2^7 .......

共有n项

这是的公式就为 若n%2==0     T(n)=T(n/2)+T(n/2)*2^(n/2);

若n%2==1     T(n)=T(n/2)+T(n/2)*2^(n/2)+ 2^n;

对于此题来讲 先把所给的循环位上的和求出来，做为基底，然后利用快速幂上面的公式求解接可以了

因为比如说12507 12507  和为2^2+2^3+ 2^2*2^len+2^3*2^len == (2^2+2^3)+(2^2+2^3)*len;这样的话就可以直接把一个序列的和all当作基底来处理了...

（不过这个公式没有找到证明啊）：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<stdlib.h>
using namespace std;
const __int64 mod=1000000007;
__int64 all;
int Pow(__int64 n,int num)      //快速幂求num^n
{if(n==0)return 1;if(n==1)return num;__int64tmp=Pow(n/2,num)%mod;tmp=tmp*tmp%mod;if(n%2==1)tmp=tmp*num%mod;return (int)tmp;
}
__int64 len;
int q;
int fun(int n)
{if(n==1)return all;__int64 tmp=fun(n/2);tmp = (tmp+tmp*Pow(n/2,q))%mod;if(n%2==1)tmp = ( tmp+Pow((n-1)*len,2)*all%mod )%mod;return (int)tmp;}char a[2000000];
int main()
{int n,ans;while(scanf("%s",a)!=EOF){cin>>n;ans=0;len=strlen(a);q=Pow(len,2)%mod;all=0;for(int i=0; i<len; i++){if(a[i]=='0'||a[i]=='5'){all=(all+Pow(i,2))%mod;}}ans=fun(n);cout<<ans<<endl;}return 0;
}

D. Block Tower

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
struct xl
{int x,y;char z;
} q[5000005];
char a[505][505];
int vis[505][505];
int dx[4]= {1,0,-1,0};
int dy[4]= {0,-1,0,1};
int n,m,i,j,len,k;
void dfs(int ans,int num)//暴搜
{int i;len++;vis[ans][num]=1;for(i=0; i<4; i++){if(a[ans+dx[i]][num+dy[i]]=='.'&&vis[ans+dx[i]][num+dy[i]]==0){dfs(ans+dx[i],num+dy[i]);}}if(len > 1)//判断是否建红色建筑{q[k].x=ans;q[k].y=num;q[k].z='D';k++;q[k].x=ans;q[k].y=num;q[k].z='R';k++;a[ans][num]='R';}len--;
}
int main()
{char c;scanf("%d%d",&n,&m);memset(vis,0,sizeof(vis));for(i=1; i<=n; ++i){c=getchar();for(j=1; j<=m; ++j)scanf("%c",&a[i][j]);}k = 0;for(i = 1; i <= n; ++i){for(j = 1; j <= m; ++j){if(a[i][j]=='.')//建蓝色建筑{q[k].x=i;q[k].y=j;q[k].z='B';k++;}}}for(i=1; i<=n; ++i){for(j=1; j<=m; ++j){if(vis[i][j]==0&&a[i][j]=='.'){len=0;dfs(i,j);}}}printf("%d\n",k);for(i=0; i<k; ++i){printf("%c %d %d\n",q[i].z,q[i].x,q[i].y);}return 0;
}

#include<iostream>
#include<cstdio>
#define N 1000000007
#define maxn 1<<25
using namespace std;
int n,k;
int a[25],b[3],dp[maxn],sum[maxn];
int DP()
{dp[0]=1;//只能从0开始for(int i=1;i<(1<<n);i++){//求出当前状态所在的位置sum[i]=0;int p;for(int j=0;j<n;j++){if(i&(1<<j)){p=j;//从小到大找第一个为1的位置break;}}sum[i]=sum[i^(1<<p)]+a[p];//去掉这个位置的状态一定已经访问过//判断当前位置是否可以停留int flag=0;for(int j=0;j<k;j++){if(sum[i]==b[j]){flag=1;break;}}//不可以的话，直接忽略，当前的路径和为0if(flag==1) continue;//动态转移for(int j=0;j<n;j++){if(i&(1<<j)){dp[i]+=dp[i-(1<<j)];if(dp[i]>N)//这里没用%dp[i]-=N;}}}return dp[(1<<n)-1];
}
int main()
{scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&a[i]);}scanf("%d",&k);for(int i=0;i<k;i++) scanf("%d",&b[i]);printf("%d\n",DP());return 0;
}