Revenge of GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2140    Accepted Submission(s): 596

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

Source

BestCoder Round #10

被坑惨了。。GCD的参数传的INT。。求出最大公约数然后求最大公约数的第K大因子就好。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iostream>
using namespace std;
typedef long long LL;
LL gcd(LL a,LL b){
    return b==0?a:gcd(b,a%b);
}
LL p[10005];
LL cmp(LL a,LL b){
    return a>b;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        LL a,b,k;
        scanf("%lld%lld%lld",&a,&b,&k);
        LL d = gcd(a,b);
        int id = 0;
        for(LL i=1;i*i<=d;i++){ ///筛选出所有因子
            if(d%i==0){
                 if(i*i==d) p[id++]=i;
                 else{
                    p[id++]=i;
                    p[id++]=d/i;
                 }
            }
        }
        if(id<k){
            printf("-1\n");
        }
        else{
            sort(p,p+id,cmp);
            printf("%lld\n",p[k-1]);
        }
    }
    return 0;
}