time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t=ssss… For example, if s= 10010, then t= 100101001010010…

Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt0,q−cnt1,q, where cnt0,q is the number of occurrences of 0 in q, and cnt1,q is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.

A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string “abcd” has 5 prefixes: empty string, “a”, “ab”, “abc” and “abcd”.

Input
The first line contains the single integer T (1≤T≤100) — the number of test cases.

Next 2T lines contain descriptions of test cases — two lines per test case. The first line contains two integers n and x (1≤n≤105, −109≤x≤109) — the length of string s and the desired balance, respectively.

The second line contains the binary string s (|s|=n, si∈{0,1}).

It’s guaranteed that the total sum of n doesn’t exceed 105.

Output
Print T integers — one per test case. For each test case print the number of prefixes or −1 if there is an infinite number of such prefixes.

Example
inputCopy
4
6 10
010010
5 3
10101
1 0
0
2 0
01
outputCopy
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.

题意看不懂，写这道题的时候卡了好久，以后做这种有特判的题目要冷静思考，把类别分开来，一类归一类。

思路：
求前缀和f[i]从0到n为0的个数减去1的个数。
inf特判：存在(f[i]<mark>x && base</mark>0)
其他：判断差值是否能整除

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MAXN = 1e5 + 5;

ll f[MAXN];
ll p[MAXN];

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin>>t;
	while(t--){
		ll n,x;
		cin >> n >> x;
		string s;
		cin >> s;
		for(int i = 0;i < n;i++){
			if (s[i] == '0') f[i+1] = f[i] + 1;
			else f[i+1] = f[i] - 1;
		}
		ll base = f[n];
		bool inf = 0;
		ll ans = 0;
		for(int i = 0;i <= n;i++){
			if(f[i]==x) ans++;
			if(f[i]==x && base==0) inf=1;
			if(i!=0 && f[i]<x && base>0 && (x-f[i])%base==0) ans++;
			if(i!=0 && f[i]>x && base<0 && (f[i]-x)%(-base)==0) ans++;			
		}
		if(inf) cout<<-1<<endl;
		else cout<<ans<<endl;
	}
	return 0;
}