Revenge of GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

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

本题大意：求出俩数的第k大的约数。

本题思想：我做这题的思想是先求出本题的最大公约数，再求出最大公约数所有的因子，用STL中set容器存储。使用反向迭代器，可求出第k大的因子。

代码如下：

#include <iostream>
#include <string>
#include <cmath>
#include <cstring>
#include <set>
typedef __int64 ll;
using namespace std;

set<ll> s;
ll GCD(ll xx,ll yy)
{
    ll r=1;
    while(r!=0)
    {
        r=xx%yy;
        xx=yy;
        yy=r;
    }
    return xx;
}

int main()
{
    int T;
    ll x,y,k,gcd,count;
    cin>>T;
    while(T--)
    {
        count=0;
        cin>>x>>y>>k;
        gcd=GCD(x,y);
        for(int i=1;i<=sqrt(gcd);i++)
        {
            if(gcd%i==0)
            {
                s.insert(i);
                s.insert(gcd/i);
            }
        }
        set<ll>::reverse_iterator rit;
        for(rit=s.rbegin();rit!=s.rend();rit++)
        {
            //cout<<*rit<<" ";
            count++;
            if(count==k)
            {
              cout<<*rit<<endl;
              break;
            }
        }
        if(count<k)cout<<"-1"<<endl;
        s.clear();
    }
    return 0;
}

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