http://codeforces.com/problemset/problem/1295/B
B. Infinite Prefixes
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.
计算前缀和后,辨别前缀和是否为0,如果是0,只需要查看输入的x是否在前缀和数组里出现过,如果出现过,则这个序列就有无穷个,不然就0个;前缀和如果非0,那么对每一位,看x减去当前位置前缀和后是否是总前缀和的整数倍,切倍数大于等于0,成立则res++;
重点就在于前缀和非0,res计算,这个方法十分巧妙,对循环节的利用非常到位,以后再遇见类似的题可以想一想这个巧点。
#include
#include
#include
using namespace std;
const int maxn=1e5+5;
char str[maxn];
int pre[maxn];
int main() {
int t;
cin>>t;
while(t--) {
int n,m;
cin>>n>>m;
scanf("%s",str+1);
pre[0]=0;
// cout<=0) res++;
}
printf("%d\n",res);
}
else {
int res=0;
for(int i=1;i<=n;i++) {
if(pre[i]==m) {
res++;
}
}
if(res) {
printf("-1\n");
}
else {
printf("0\n");
}
}
}
}
//6 353709959
//110000