codefroces 873 B. Balanced Substring && X73(前缀和思想）

Posted 2020-10-10 十年换你一句好久不见

B. Balanced Substring

You are given a string s consisting only of characters 0 and 1. A substring [l,?r] of s is a string s_ls_l?+?1s_l?+?2... s_r, 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.

前缀和离散化

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <cstdlib>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
char s[100005];
int a[200005],n,ans=0,val=0;
int main()
{
    scanf("%d%s",&n,&s);
    memset(a,0,sizeof(a));
    for(int i=0;i<n;i++)
    {
        ans+=s[i]==‘0‘?-1:1;
        if(ans==0) val=i+1;
        else
        {
            if(!a[ans+n]) a[ans+n]=i+1;
            else val=max(val,i-a[ans+n]+1);//若中间值存在必为0,因此左右指针对应的值相等
        }
    }
    printf("%d\n",val);
    return 0;
}

给你一个数组，求数组中连续区间内和为73的区间个数

输入n(0<n<=100000),接下来输入n个数a_i(0<a_i<75);

输出

连续区间内和为73的区间个数，若没有输出"X"

前缀和维护数组。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <cstdlib>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll val[100005],n,ans=0,k=0;
map<ll,ll>a;
int main()
{
    scanf("%lld",&n);
    a[0]=1;
    for(int i=1;i<=n;i++) scanf("%lld",&val[i]);
    for(int i=1;i<=n;i++)
    {
        ans+=val[i];
        a[ans]++;
        if(ans>=73) k+=a[ans-73];
    }
    if(k) printf("%lld\n",k);
    else printf("X\n");
    return 0;
}