You are given string ss of length nn consisting of 0-s and 1-s. You build an infinite string tt as a concatenation of an infinite number of strings ss, or t=ssss…t=ssss… For example, if s=s= 10010, then t=t= 100101001010010...
Calculate the number of prefixes of tt with balance equal to xx. The balance of some string qq is equal to cnt0,q−cnt1,qcnt0,q−cnt1,q, where cnt0,qcnt0,q is the number of occurrences of 0 in qq, and cnt1,qcnt1,q is the number of occurrences of 1 in qq. 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 TT (1≤T≤1001≤T≤100) — the number of test cases.
Next 2T2T lines contain descriptions of test cases — two lines per test case. The first line contains two integers nn and xx (1≤n≤1051≤n≤105, −109≤x≤109−109≤x≤109) — the length of string ss and the desired balance, respectively.
The second line contains the binary string ss (|s|=n|s|=n, si∈{0,1}si∈{0,1}).
It‘s guaranteed that the total sum of nn doesn‘t exceed 105105.
Output
Print TT integers — one per test case. For each test case print the number of prefixes or −1−1 if there is an infinite number of such prefixes.
Example
input
Copy
4
6 10
010010
5 3
10101
1 0
0
2 0
01
output
Copy
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of tt: with length 2828, 3030 and 3232.
题解： 前缀和记录字符串每个位置的平衡值。 注意平衡值x可能是负的
#include<iostream>
#include<string>
#include <cstdlib>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include <iomanip>
// #pragma comment(linker, "/STACK:1024000000,1024000000")
// #define pi acos(-1)
// #include<bits/stdc++.h>usingnamespace std;
typedef longlong ll;
#define INF 0x7f7f7f7f //2139062143#define INF1 0x3f3f3f3f //1061109567#define INF2 2147483647
#define llINF 9223372036854775807
#define pi 3.141592653589793//23846264338327950254#define pb push_back
#define ll long long
#define debug cout << "debug\n";
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);#define CNM ios::sync_with_stdio(false);cin.tie(NULL);
#define scai(x) scanf("%d", &x)
#define sca2i(x, y) scanf("%d %d", &x, &y)
#define scaf(x) scanf("%lf", &x)
#define sca2f(x, y) scanf("%lf %lf", &x, &y)
#define For(m,n) for (int i = m; i < n; i++)
inline int read(){
int s=0,w=1;
char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)w=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘) s=s*10+ch-‘0‘,ch=getchar();
return s*w;
}
#define local
#ifdef local
#endif#define MAX 10233
#define LCH(i) ((i) << 1)
#define RCH(i) ((i) << 1 | 1)
constint N = 1e5 + 5;
ll a[N] = {0};
int main()
{
freopen("in.txt","r",stdin);
CNM;
ll t;
cin >> t;
while (t--)
{
ll n, v;
cin >> n >> v;
string s;
cin >> s;
ll ans = 0;
for (int i = 0; i < n; i++)
{
a[i + 1] = a[i] + (s[i] == ‘0‘?1:-1);
}
for (int i = 0; i < n; i++)
{
int dv = v - a[i];
if (a[n] == 0)
{
if (dv == 0)
ans = -1;
}
elseif (dv % a[n] == 0)
{
dv /= a[n];
if (dv >= 0) ans++;
}
}
printf("%lld\n", ans);
}
}
