Infinite Prefixes
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given string s of length nn 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…t=ssss… For example, if s= 10010, then 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≤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 ( ) — the length of string s and the desired balance, respectively.
The second line contains the binary string ss (|s|=n, si∈{0,1}).
It's guaranteed that the total sum of nn doesn't exceed .
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
input
4 6 10 010010 5 3 10101 1 0 0 2 0 01
output
3 0 1 -1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.
题目大意
给一个长度已知可以无限重复的01串s，定义一个字符串的q值为’0’字符的个数减去’1’字符的个数，问给定字符串s的所有前缀中（包括空字符串）q值为x的个数
题目不难，但是细节比较麻烦，无非就是一个周期问题，但是一个周期之内会有极大极小，就可能导致多解，首先预处理出串的每一个位置的q值，然后遍历一遍01串，如果（x-该位置的q值）%（01串最后一个位置q值）=0，还没完，因为你得检验如果该位置+（x-该位置的q值）/（01串最后一个位置q值）>0就ok,完全可以，当然想到这还不够，，还有无限解的情况
何时解无穷？
01串最后一个位置q值=0： 如果x=0，那么一定有无穷解；
如果x0，那么遍历一遍字符串，要是某一个位置q值=x，则也有无穷解
当然x=0的时候一定有解，因为空前缀也行~
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rg register ll
ll a[100005];
int main()
{
ll t;
cin>>t;
for(rg i=1;i<=t;i++)
{
ll n,x,ans=0;
cin>>n>>x;
string s;
cin>>s;
s[0]=='0'?a[0]=1:a[0]=-1;
for(rg i=1;s[i];i++)
{
a[i]=a[i-1];
if(s[i]=='0')a[i]++;
else a[i]--;
}
if(!x&&!a[n-1])
{
cout<<-1<<endl;
continue;
}
if(!x)ans++;
if(!a[n-1])
{
ll tep=0,flag=0;
for(rg i=0;i<n;i++)
{
if(a[i]==x)
{
cout<<-1<<endl;
flag=1;
break;
}
}
if(!flag)cout<<0<<endl;
continue;
}
for(rg i=0;i<n;i++)
{
if((x-a[i])%a[n-1]==0&&i+1+(x-a[i])/a[n-1]*n>0)ans++;
}
cout<<ans<<endl;
}
while(1)getchar();
return 0;
}