2016-2017 National Taiwan University World Final Team Selection Contest J - Zero Game

题目：

You are given one string S consisting of only '0' and '1'. You are bored, so you start to play with the string. In each operation, you can move any character of this string to some other position in the string. For example, suppose . Then you can move the first zero to the tail, and S will become '0100'.

Additionally, you have Q numbers K₁, K₂, ..., K_Q. For each i, you wonder what can be the maximum number of consecutive zeroes in the string if you start with S and use at most K_i operations. In order to satisfy your curiosity, please write a program which will find the answers for you.

Input

The first line of input contains one string S. The second line of input contains one integer Q. Each of the following Q lines contains one integer K_i indicating the maximum number of operations in i-th query.

• 2 ≤ N ≤ 10⁶
• the length of S is exactly N characters
• S consists of only '0' and '1'
• 1 ≤ Q ≤ 10⁵
• N × Q ≤ 2 × 10⁷
• 1 ≤ K_i ≤ 10⁶

Output

For each query, output one line containing one number: the answer for this query.

Example

Input

0000110000111110
5
1
2
3
4
5

Output

5
8
9
9
9

思路：
　　对于每个区间[l,r],如果sum[r]-sum[l-1]<=k，则说明可以区间内所有1踢掉，然后还可以加入k-(sum[r]-sum[l1])个0进来。
　　所以合法区间的贡献为(2*sum[l-1]-l)+(r-2*sum[r])+k+1.
　　将判断合法的等式变形，可以得到：sum[r]-k<=sum[l-1]
　　可以看出合法的r单调，所以可以枚举l，用单调队列维护答案。

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
 #define PB push_back
 typedef long long LL;
 typedef pair<int,int> PII;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int K=1e6+;
 const int mod=1e9+;

 int n,m,k,sum[K],q[K];
 char ss[K];
 int main(void)
 {
     scanf("%s",ss+);
     n=strlen(ss+);
     for(int i=;i<=n;i++)   sum[i]=sum[i-]+(ss[i]==''?:);
     scanf("%d",&m);
     while(m--)
     {
         int k,ans=,st=,se=;
         scanf("%d",&k);
         for(int i=,j=;i<=n;i++)
         {
             while(j<=n&&sum[j]-k<=sum[i-])
             {
                 while(st>se&&q[st]-*sum[q[st]]<=j-*sum[j]) st--;
                 q[++st]=j++;
             }
             while(st>se&&q[se+]<i) se++;
             ans=max(ans,*sum[i-]-i+q[se+]-*sum[q[se+]]+k+);
         }
         ans=max(,ans);
         ans=min(ans,n-sum[n]);
         printf("%d\n",ans);
     }
     return ;
 }