牛牛的方程式

更好的阅读体验

裴蜀定理的推广，若，则，负数可以转化为正数处理

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

#define int long long

using namespace std;

const int N = 1e6 + 10, M = 2e6 + 10, inf = 0x3f3f3f3f;

inline int read() {
    int res = 0; char ch = getchar();
    while (!isdigit(ch)) ch = getchar();
    while (isdigit(ch)) res = (res << 3) + (res << 1) + (ch ^ 48), ch = getchar();
    return res;
}

int n, m, cnt, a[4], d;

int gcd(int a, int b) {return b ? gcd(b, a % b) : a;}

signed main() {
    int T = read();
    while (T--) {
        a[1] = read(); a[2] = read(); a[3] = read(); d = read();
        if (a[1] == 0 && a[2] == 0 && a[3] == 0) {
            printf(d ? "NO\n" : "YES\n"); continue;
        }
        if (d == 0) {printf("YES\n"); continue;}
        int p = 1, c = a[p]; while(!c && p <= 3) c = a[++p];
        for (int i = p; i <= 3; i++) {
            if (a[i]) c = gcd(c, a[i]);
        }
        if (c > d) {printf("NO\n"); continue;}
        if (gcd(c, d) != c) {printf("NO\n"); continue;}
        printf("YES\n");
    }
    return 0;
}