CF1200C

CF1200C

Meaning of the questions:

Q are divided into inner and outer circle n, parts of m, each numeral has asked if can go from one part to another, there must be a 12 o'clock line.

solution:

If there is a through-wall 1 and 2, so that the sides will not communicate. Then at n or m / common denominator of such walls will clearly appear.

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std; 

#define LL long long 

LL n,m,q;

inline LL gcd(LL a,LL b) {
    return !b ? a : gcd(b,a%b);
}
 
int main() { 
    scanf("%lld%lld%lld",&n,&m,&q); 
    LL G = gcd(n, m); 
    LL nn = n / G, mm = m / G; 
    while (q--) {
        LL x1, y1, x2, y2; 
        scanf("%lld%lld%lld%lld",&x1,&y1,&x2,&y2); 
        y1--, y2--; 
        LL ans1, ans2; 
        if(x1 == 1) ans1 = y1 / nn; 
        else ans1 = y1 / mm; 
        if(x2 == 1) ans2 = y2 / nn; 
        else ans2 = y2 / mm; 
        if(ans1 == ans2) puts("YES"); 
        else puts("NO"); 
    }
    //system("pause");
    return 0; 
}