2017 ACM Arabella Collegiate Programming Contest Sherlock Bones

Sherlock Bones

time limit per test

1.5 s

memory limit per test

256 MB

input

standard input

output

standard output

The great dog detective Sherlock Bones is on the verge of a new discovery. But for this problem, he needs the help of his most trusted advisor -you- to help him fetch the answer to this case.

He is given a string of zeros and ones and length N.

Let F(x, y) equal to the number of ones in the string between indices x and y inclusively.

Your task is to help Sherlock Bones find the number of ways to choose indices (i, j, k) such that i < j < k, sj is equal to 1, and F(i, j) is equal to F(j, k).

Input

The first line of input is T – the number of test cases.

The first line of each test case is an integer N (3 ≤ N ≤ 2 × 105).

The second line is a string of zeros and ones of length N.

Output

For each test case, output a line containing a single integer- the number of ways to choose indices (i, j, k).

Example

input

3
5
01010
6
101001
7
1101011

output

2
3
7

题意：给你一个长度为n的01字符串，问能找出多少个三元组(i,j,k)满足ch[j]==1&&i<j<k&&F(i,j)==F(j,k)(F(x,y)表示区间[x,y]中字符1的个数)

解题思路：可以先将问题转换为有多少个区间中1的个数为奇数，这个可以dp解决，然后减去字符1在最左端，右边全为0的和字符在最右端，左边全为1的即可

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <string>  
#include <algorithm>  
#include <cmath>  
#include <map>  
#include <set>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <bitset>  
#include <functional>  

using namespace std;

#define LL long long  
const int INF = 0x3f3f3f3f;
const int maxn = 200000 + 10;

char ch[maxn];
int L[maxn], R[maxn], n;
long long dp[maxn][2];

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		scanf("%s", ch + 1);
		L[0] = R[n + 1] = dp[0][0] = dp[0][1] = 0;
		for (int i = 1; i <= n; i++)
		{
			if (i > 1 && ch[i - 1] == '1') L[i] = 1;
			else L[i] = L[i - 1] + 1;
		}
		for (int i = n; i >= 1; i--)
		{
			if (i < n && ch[i + 1] == '1') R[i] = 1;
			else R[i] = R[i + 1] + 1;
		}
		LL ans = 0;
		for (int i = 1; i <= n; i++)
		{
			if (ch[i] == '1')
			{
				dp[i][0] = dp[i - 1][1];
				dp[i][1] = dp[i - 1][0] + 1;
			}
			else
			{
				dp[i][0] = dp[i - 1][0] + 1;
				dp[i][1] = dp[i - 1][1];
			}
			ans = ans + dp[i][1];
		}
		for (int i = 1; i <= n; i++)
		{
			if (ch[i] == '1')
			{
				ans = ans - (L[i] - 1);
				ans = ans - (R[i] - 1);
				ans--;
			}
		}
		printf("%lld\n", ans);
	}
	return 0;
}