Fraction Construction Problem

思路

$\frac{c}{d}-\frac{e}{f}=\frac{a}{b}$

$a<b\&f<b$

$1\le 1,e\le 4\times 10^{12}$

• 当b = 1时，一定无解。

• gcd(a, b) != 1，能够较为简单地构造出来。

• $b=p^k$，我们设$d=p^{k-x},f=p^x$，左边方程通分得到$\frac{c{p}^x-e{p}^{k-x}}{p^k}$，则方程$c{p}^x-e{p}^{k-x}=a$有解，那么一定有$gcd(p^x,p^{k-x})\mid a$，

  只能$x=0$,所以$d=p^k=b$不符合要求。

• 所以这时$b$一定有至少两个质因，我们不妨设$d\times f=b$，$d,f$互质，这样我们只要通过拓展欧几里德去求解即可。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f;
const double eps = 1e-7;const int N = 1e4 + 10;int prime[N], cnt;ll c, d, e, f;bool st[N];void init() {for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;}}
}bool get_fac(ll n) {for(int i = 0; prime[i] * prime[i] <= n; i++) {if(n % prime[i] == 0) {d = 1;while(n % prime[i] == 0) {n /= prime[i];d *= prime[i];}if(n == 1) return false;return true;}}return false;
}ll exgcd(ll a, ll b, ll & x, ll & y) {if(!b) {x = 1, y = 0;return a;}ll d = exgcd(b, a % b, x, y);ll temp = x;x = y;y = temp - a / b * y;return d;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;init();scanf("%d", &T);while(T--) {ll a, b;scanf("%lld %lld", &a, &b);int gcd = __gcd(a, b);if(gcd != 1) {printf("%lld %lld %lld %lld\n", a / gcd + 1, b / gcd, 1, b / gcd);continue;}if(!get_fac(b)) {puts("-1 -1 -1 -1");continue;}f = b / d;if(f < d) swap(f, d);ll x, y;ll g = exgcd(f, d, x, y);while(y > 0) x += d, y -= f;printf("%lld %lld %lld %lld\n", x * a, d, -y * a, f);}return 0;
}