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.
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.
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
8 11010111
4
3 111
0
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.
题解：a[i]表示的是[0..i]区间内1和0的个数之差。
如果a[i]==a[j]的话，那么[i+1,j]区间1和0的数量相同。
确定下每个a[i]出现的最早下标和最晚下标，做差一下，然后更新答案即可。
代码：
#include <bits/stdc++.h>
using namespace std;
int n,x,k;
const int maxn = 100010;
int a[maxn];
map<int,int> mark;
int main(){cin>>n;int nums = 0;for(int i = 1;i <= n;++i){char c;scanf(" %c",&c);if(c == '0') nums--;else nums++;a[i] = nums;if(!mark[nums]) mark[nums] = i;}mark[0] = 0;int ans = 0;for(int i = 1;i <= n;++i){ans = max(ans,i - mark[a[i]]);}cout<<ans<<endl;
}
)
