zoukankan      html  css  js  c++  java
  • 2016-2017 National Taiwan University World Final Team Selection Contest J

    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, ..., KQ. 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 Ki 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 Ki 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 ≤ Ki ≤ 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。

     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 
     5 #define MP make_pair
     6 #define PB push_back
     7 typedef long long LL;
     8 typedef pair<int,int> PII;
     9 const double eps=1e-6;
    10 const double pi=acos(-1.0);
    11 const int K=1e6+7;
    12 const int mod=1e9+7;
    13 
    14 int n,m,k,sum[K],q[K];
    15 char ss[K];
    16 int main(void)
    17 {
    18     scanf("%s",ss+1);
    19     n=strlen(ss+1);
    20     for(int i=1;i<=n;i++)   sum[i]=sum[i-1]+(ss[i]=='1'?1:0);
    21     scanf("%d",&m);
    22     while(m--)
    23     {
    24         int k,ans=0,st=0,se=0;
    25         scanf("%d",&k);
    26         for(int i=1,j=1;i<=n;i++)
    27         {
    28             while(j<=n&&sum[j]-k<=sum[i-1])
    29             {
    30                 while(st>se&&q[st]-2*sum[q[st]]<=j-2*sum[j]) st--;
    31                 q[++st]=j++;
    32             }
    33             while(st>se&&q[se+1]<i) se++;
    34             ans=max(ans,2*sum[i-1]-i+q[se+1]-2*sum[q[se+1]]+k+1);
    35         }
    36         ans=max(0,ans);
    37         ans=min(ans,n-sum[n]);
    38         printf("%d
    ",ans);
    39     }
    40     return 0;
    41 }
    
    
  • 相关阅读:
    ssdb使用笔记
    跟我学爬虫-2-使用正则表达式解析文本
    跟我学爬虫-1-爬虫简介
    python int函数转换浮点型字符串的坑???
    python使用smtplib和email发送腾讯企业邮箱邮件
    php文件之间传值的三种主流并且常用的方式
    验证码的输入框与图片不能对齐问题
    web前端命名规范
    css盒子模型
    css基础
  • 原文地址:https://www.cnblogs.com/weeping/p/7301661.html
Copyright © 2011-2022 走看看