这道题是为了记录一下这个差点被忘记，但是十分重要的数学结论：
gcd(X,Y) = gcd(X, Y - K*X) = gcd(X - K * Y, Y)
#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; //gcd(x, y) = gcd(x, y - k *x) = gcd(x - k * y, y) //其实这个就是欧几里得算法求GCD的主要思想 typedef long long ll; ll gcd(ll n, ll m) { if(n % m == 0) return m; return gcd(m, n % m); } int main() { int CASE; scanf("%d", &CASE); for(int cas = 1; cas <= CASE; ++cas) { ll x, y; scanf("%lld%lld", &x, &y); int ans = 1; while(x != y) { if(x > y) swap(x, y); ll pre = y; ll g = gcd(x, y); y = y / (1 + x / g); if(y + x * y / gcd(x, y) != pre) break; ans += 1; } printf("Case #%d: %d\n", cas, ans); } return 0; }
