题解 | 仰望水面的歪

#include <iostream>
#include <vector>
using namespace std;

long gcd(long a, long b){
    return b==0? a : gcd(b,a%b);
}

int main() {
    long  n, h;
    cin >> n >> h;
    vector<vector<long>> vec(n, vector<long>(3));
    for(int i=0; i<n; ++i){
        cin >> vec[i][0] >> vec[i][1] >> vec[i][2];
    }
    for(int i=0; i<n; ++i){
        long a = vec[i][0], b = vec[i][1], c =  h-vec[i][2]+h;
        long k = gcd(gcd(a, b), c);
        cout << a/k << ' ' << b/k << ' ' << c/k << endl;
    }

    return 0;
}
// 64 位输出请用 printf("%lld")