【HDU

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 <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>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;}