Infinite Prefixes CodeForces - 1295B
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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".
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.
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.
4 6 10 010010 5 3 10101 1 0 0 2 0 01
3 0 1 -1
In the first test case, there are 3 good prefixes of tt: with length 2828, 3030 and 3232.
#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> using namespace std; typedef long long 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 "; // 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) const int 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; } else if (dv % a[n] == 0) { dv /= a[n]; if (dv >= 0) ans++; } } printf("%lld ", ans); } }
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