Codeforces 1398C- Good Subarrays（区间值为0的个数变形-思维）

题目链接：https://codeforces.com/problemset/problem/1398/C
博客园食用链接： https://www.cnblogs.com/lonely-wind-/p/13509265.html

You are given an array $a_1,a_2,\dots ,a_n$ consisting of integers from 0 to 9. A subarray $a_l,a_{l+1},a_{l+2},\dots ,a_{r-1}$,ar is good if the sum of elements of this subarray is equal to the length of this subarray ${\sum }_{i=l}^r{a}_i=r-l+1$.

For example, if a=[1,2,0], then there are 3 good subarrays: $a_{1\dots 1}=[1],a_{2\dots 3}=[2,0]$ and $a_{1\dots 3}=[1,2,0]$.

Calculate the number of good subarrays of the array a.

Input
The first line contains one integer t (1≤t≤1000) — the number of test cases.

The first line of each test case contains one integer $n(1\le n\le 10^5)$ — the length of the array a.

The second line of each test case contains a string consisting of n decimal digits, where the i-th digit is equal to the value of $a_i$.

It is guaranteed that the sum of n over all test cases does not exceed $10^5$.

Output
For each test case print one integer — the number of good subarrays of the array a.

input
3
3
120
5
11011
6
600005
output
3
6
1

Note
The first test case is considered in the statement.

In the second test case, there are 6 good subarrays:$a_{1\dots 1},a_{2\dots 2},a_{1\dots 2},a_{4\dots 4},a_{5\dots 5}$ and $a_{4\dots 5}$.

In the third test case there is only one good subarray: $a_{2\dots 6}$.

题目大意：给你一个字符串序列，你需要找到有多少个区间使得其区间内每个元素的和为区间长度。

emmm，对于其公式我们很容易看出其区间的平均值为1，我们将其每个数-1，就可变为另一个东西：求区间和值为0的有多少个。那么我们只需要记录其每个前缀和出现的次数就好了，也就是说对于一个前缀和，如果它第二次出现了，那么可以说明中间一定有存在构成一个0的区间。那么我们只需要将其每次出现的次数加上就好了。

以下是AC代码：

#include 
using namespace std;

typedef long long ll;
const int mac=1e5+10;

char s[mac];
int a[mac],sum[mac];

int main(int argc, char const *argv[])
{
	int t;
	scanf ("%d",&t);
	while (t--){
		int n;
		scanf ("%d",&n);
		scanf ("%s",s+1);
		for (int i=1; i<=n; i++)
			a[i]=s[i]-'0'-1,sum[i]=sum[i-1]+a[i];
		unordered_map<int,int>mp;
		ll ans=0;
		mp[0]=1;
		for (int i=1; i<=n; i++){
			if (mp[sum[i]]) ans+=mp[sum[i]],mp[sum[i]]++;
			else mp[sum[i]]++;
		}
		printf ("%lld\n",ans);
	}
	return 0;
}