牛客多校__Crazy Binary String

鏈接：https://ac.nowcoder.com/acm/contest/883/B
來源：牛客網
 

時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld

題目描述

ZYB loves binary strings (strings that only contains `0' and `1'). And he loves equal binary strings\textit{equal binary strings}equal binary strings more, where the number of `0' and the number of `1' in the string are equal.

ZYB wants to choose a substring from an original string  T\ T T so that it is an equal binary string\textit{equal binary string}equal binary string with the longest length possible. He also wants to choose a subsequence of  T\ T T which meets the same requirements.

A string  v\ v v is a substring of a string  w\ w w if  v\ v v is empty, or there are two integers  l\ l l and r (1≤l≤r≤∣w∣)r \ (1 \le l \le r \le |w|)r (1≤l≤r≤∣w∣) such that v=wlwl+1⋯wrv=w_lw_{l+1}\cdots w_rv=wl​wl+1​⋯wr​. A string  v\ v v is a subsequence of a string  w\ w w  if it can be derived from  w\ w w  by deleting any number (including zero) of characters without changing the order of the remaining characters. 

For simplicity, you only need to output the maximum possible length. Note that the empty string is both a substring and a subsequence of any string.

輸入描述:

The first line of the input contains a single integer N (1≤N≤100000)N \ (1 \le N \leq 100000)N (1≤N≤100000), the length of the original string  T\ T T. The second line contains a binary string with exactly  N\ N N characters, the original string  T\ T T.

輸出描述:

Print two integers  A\ A A and  B\ B B, denoting the answer for substring and subsequence respectively.

示例1

輸入

複製

8
01001001

輸出

複製

4 6

弄個前綴和，然後出現相同前綴的，把座標相減去最大

#include <bits/stdc++.h>

using namespace std;

const int maxn = 3e5+9;

char s[maxn];
int a[maxn];

int main(){
    int n;
    cin>>n;
    int len=0;
    scanf("%s",s+1);
    memset(a,-1,sizeof(a));
    int sum=100000;
    a[100000]=0;
    int cnt=0;
    for(int i=1;s[i];i++){
        if(s[i]=='0'){
            cnt++;
            sum--;
            if(a[sum]==-1)a[sum]=i;
        }
        else{
            sum++;
            if(a[sum]==-1)a[sum]=i;
        }
        len=max(len,i-a[sum]);
    }
    printf("%d %d\n",len,min(cnt,n-cnt)*2);
    return 0;
}