Educational Codeforces Round 81 (Rated for Div. 2) B题

Infinite Prefixes

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.

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思维题

记录字符串s每个位置的前缀，也就是01之差的前缀和，只要枚举每个位置是否可以做为答案就行，有点巧妙；

代码：

#include<bits/stdc++.h>
#define LL long long
#define pa pair<int,int>
#define lson k<<1
#define rson k<<1|1
//ios::sync_with_stdio(false);
#define inf 0x3f3f3f3f
using namespace std;
const int N=100100;
const int M=1000100;
const LL mod=20010905;
int sum[N],sum1[N],sum2[N];//正负 
int main(){
	ios::sync_with_stdio(false);
	int t;
	cin>>t;
	while(t--){
		memset(sum1,0,sizeof(sum1));
		memset(sum2,0,sizeof(sum2));
		int n,x;
		string s;
		cin>>n>>x;
		cin>>s;
		int y=0;
		for(int i=0;i<n;i++){
			if(s[i]=='0') y++;
			if(s[i]=='1') y--;
			if(y>=0) sum1[y]++;
			else sum2[-y]++;
			sum[i]=y;
		}
		if(!y){
			if(x>100000||x<-100000) cout<<0<<endl;
			else if(x>=0){
				if(sum1[x]) cout<<-1<<endl;
				else cout<<0<<endl;
			}
			else if(x<0){
				if(sum2[-x]) cout<<-1<<endl;
				else cout<<0<<endl;
			}
		}
		else{
			int ans=0;
			if(x==0) ans++;
			for(int i=0;i<n;i++){
				if((x-sum[i])/y>=0&&!((x-sum[i])%y)) ans++;
			}
			cout<<ans<<endl;
		}
	}
	return 0;
}