Infinite Prefixes
time limit per test 2 seconds
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,qis 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≤10^5, −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 10^5.
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.
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.
题解
让我们将长度为i的前缀表示为pref(i)。
我们可以注意到每个pref(i)=k⋅pref(n)+ pref(i mod n)
其中k =⌊in⌋,而+是串联。
则长度为i的前缀的bal(i)等于k⋅bal(n)+ bal(i mod n)。现在有两种情况:bal(n)等于或不等于0。
如果bal(n)= 0,则如果存在j(0≤j<n),则对于每一个k≥0bal(j + kn)= x,bal(j)= then,则答案为-1。
j不超过一个可能的k:因为方程j +k⋅bal(n)= x有零个或一个解。
当且仅当x−j≡0modbal(n)并且k = x-jbal(n)≥0时,该解才存在。
因此,仅需预先计算bal(n),并针对每个0≤j<n检查方程式。
#include<bits/stdc++.h>
using namespace std;
int main()
{
long long T,n,x;
string s;//用char数组慢
scanf("%lld",&T);
while(T--)
{
cin>>n>>x>>s;
int num0=0,num1=0;
for(int i=0;i<n;i++)
{
if(s[i]=='0')num0++; else num1++;
}
int k=num0-num1;
if(k==0)
{
int i;
for(i=0;i<n&&k!=x;i++)
{
if(s[i]=='0')k++; else k--;
}
if(k==x)printf("-1\n");
else printf("0\n");
}
else
{
int ans=0,j=0;
for(int i=0;i<n;i++)
{
if((x-j)%k==0&&(x-j)/k>=0)ans++;
if(s[i]=='0')j++; else j--;
}
printf("%d\n",ans);
}
}
}