Revenge of GCD （枚举+gcd）

Revenge of GCD

 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

Output
For each test case, output the k-th GCD of X and Y. If no such integer exists, output -1.
Sample Input

3
2 3 1
2 3 2
8 16 3

Sample Output

1
-1
2

题意：

求x，y的第k大公因数

分析：

一开始还以为什么很难的题按着算术基本定理想了很久，原来是枚举gcd的因子，fuck？

首先求出最大公因数，然后在枚举出gcd的所有因子排个序就可以了

可以使用vector

注意一开始一开始把1和它本身放入，但是要判断如果它本身就是1不可以放两次！！

然后常规操作，从2开始枚举因子即可$\sqrt{g}$的复杂度，排序log复杂度，求gcd log复杂度

code:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){
    return b ? gcd(b,a%b) : a;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        ll x,y,k;
        scanf("%lld%lld%lld",&x,&y,&k);
        ll g = gcd(x,y);
        vector<ll> ans;
        ans.push_back(1LL);
        if(g != 1)
            ans.push_back(g);
        for(ll i = 2; i * i <= g; i++){
            if(g % i == 0){
                ans.push_back(i);
                if(i * i != g)
                    ans.push_back(g/i);
            }
        }
        sort(ans.begin(),ans.end());
        if(k > ans.size())
            printf("-1\n");
        else
            printf("%lld\n",ans[ans.size()-k]);
    }
    return 0;
}