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.
题意:给一个长度为n的字符串,问其子串中含有1 0个数相同的最长字串长度;
思路:将0换成-1,算出每个点的前缀和,若两个点前缀和相同,从第一个点到第二个点之前的字符串则为平衡串,求出最长的即可
AC代码:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<map>
#define maxn 1000010
using namespace std;
int n;
char str[100005];
struct node
{
int val;
int id;
}pre[100005];
bool cmp(node a,node b)
{
if(a.val==b.val) return a.id<b.id;
return a.val<b.val;
}
int main()
{
while(~scanf("%d",&n))
{
scanf("%s",str+1);
pre[1].val=0;pre[1].id=1;
for(int i=2;i<=n+1;i++)//要算到n后面一个数
{
int x=str[i-1]=='0'?-1:1;
pre[i].val=pre[i-1].val+x;
pre[i].id=i;
}
sort(pre+1,pre+2+n,cmp);//按前缀和排序
int len=0,tmp=pre[1].val;
for(int i=1;i<=n+1;i++)
{
int tmp=pre[i].val,p=pre[i].id;
while(i<=n&&pre[i+1].val==tmp) i++;
len=max(len,pre[i].id-p);
}
printf("%d\n",len);
}
return 0;
}