uva 10673 - Play with Floor and Ceil

扩展欧几里德算法：找出一对整数(x, y)，是的ax + by = gcd(a, b)，哈哈，直接套

#include <iostream>#include <cmath>using namespace std;long long x, k;long long lx, ly;void gcd(long long a, long long b, long long &d, long long &p, long long &q) {if(b == 0) {d = a;p = 1;q = 0;}else {//这算法最最最要注意这里的p,q要调换顺序gcd(b, a % b, d, q, p);q -= p * (a / b);}}//#define TESTint main() {#ifdef TESTfreopen("input.txt", "r", stdin);#endifint t;cin >> t;long long p, q, d;while(t--) {cin >> x >> k;lx = floor((double)x / (double)k);ly = ceil((double)x / (double)k);gcd(lx, ly, d, p, q);d = x / d;cout << p * d << " " << q * d << endl;}return 0;}