题解 | 仰望水面的歪

#include <iostream>
using namespace std;
#include<cmath>
int gcd(long int a, long int b) {
    if(b == 0) {
        return a;
    }
    return gcd(b, a % b);
}
int main() {
    int  a, b;
    while (cin >> a >> b) { 
        for(int i=0;i<a;i++)
        {long int   x,y,z;
        cin >>x>>y>>z;
        z=2*b-z;
        long int g=gcd(max(x,y),min(x,y));
        if(g>1)g=gcd(max(g,z),min(g,z));
            if(g>1){x/=g;y/=g;z/=g;}
        cout <<x<<' '<<y<<' '<<z<< endl;
        }
    }
}