zoukankan      html  css  js  c++  java
  • 【Cf edu 30 B. Balanced Substring】

    time limit per test 1 second

    memory limit per test 256 megabytes

    input standard input

    output standard output

    You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.

    You have to determine the length of the longest balanced substring of s.

    Input

    The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.

    The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.

    Output

    If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.

    Examples

    input

    8
    11010111

    output

    4

    input

    3
    111

    output

    0

    Note

    In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.

    In the second example it's impossible to find a non-empty balanced substring.

    【翻译】给出01串,求出最长连续子串使得01个数相同

    题解:
         ①经典思想:维护差值——将0设为-1

         ②前缀和中,前缀和相同的两点可构成合法区间,直接统计最大值就是了

    #include<stdio.h>
    #include<algorithm>
    #define go(i,a,b) for(int i=a;i<=b;i++)
    using namespace std;
    const int N=100004;
    int n,a[N],pos[N],_[N],sum[N],ans;
    int main()
    {
        scanf("%d",&n);
        go(i,1,n)scanf("%1d",a+i);
        go(i,1,n)sum[i]=sum[i-1]+(a[i]?1:-1);
        go(i,1,n)if(pos[sum[i]]||sum[i]==0)
        ans=max(ans,i-pos[sum[i]]);else pos[sum[i]]=i;
        printf("%d
    ",ans);return 0;
    }//Paul_Guderian

    .

  • 相关阅读:
    ThinkPhp学习11
    ThinkPhp学习10
    1.自我介绍
    Axure高级教程--在原型中插入视频
    Axure制作iphone手机交互模型—覆盖切换
    对产品的一些总结
    详解Axure的Masters功能
    详解使用Axure 制作Tab切换功能
    产品经理的初识
    作为产品经理--如何写好PRD文档
  • 原文地址:https://www.cnblogs.com/Damitu/p/7745321.html
Copyright © 2011-2022 走看看