HDU 5019 Revenge of GCD

Revenge of GCD

Problem Description

In mathematics, the greatest common divisor (gcd), also known as the greatest common factor (gcf), highest common factor (hcf), or greatest common measure (gcm), of two or more integers (when at least one of them is not zero), is the largest positive integer that divides the numbers without a remainder.
—Wikipedia

Today, GCD takes revenge on you. You have to figure out the k-th GCD of X and Y.

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case only contains three integers X, Y and K.

[Technical Specification]

1. 1 <= T <= 100
2. 1 <= X, Y, K <= 1 000 000 000 000

求最大公因子，刚开始被数据吓到了，没想到暴力都过得去。。。。测试数据太水了

先求出GCD（x，y），他们的其他公因子肯定是这个数的因子（用短除法求最大公因子理解下）

然后，从1到根号这个数枚举，

ll t = __gcd(x, y);
        for(int i = 1; i * i <= t; i++) {
            if(t % i == 0) {
                ans[book++] = i;
                if(i * i != t)
                    ans[book++] = t/ i;
            }
        }

#include
#include
#include
#define _for(i,a,b) for(int i=(a);i<=(b);i++)
using namespace std;
typedef long long ll;
ll x, y, k, t;
ll book;
ll ans[1000000];
int main() {
    scanf("%lld", &t);
    while(t--) {
        book = 0;
        scanf("%lld %lld %lld", &x, &y, &k);
        ll t = __gcd(x, y);
        for(ll i = 1; i * i <= t; i++) {//int 居然卡了我的时间，，，t了
            if(t % i == 0) {
                ans[book++] = i;
                if(i * i != t)
                    ans[book++] = t/ i;
            }
        }
        if(book < k)
            printf("-1\n");
        else {
            sort(ans, ans + book);
            printf("%lld\n", ans[book-k]);
        }
    }
    return 0;
}