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最新推荐文章于 2020-04-19 14:32:33 发布

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本文探讨了一道数学与编程结合的竞赛题目，目标是找出两个整数X和Y的第K大的公约数。文章分享了如何通过求最大公约数及其因子来解决这一问题的思路，并附上了AC代码实现。

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 <= T <= 100
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：给你一组数x,y,k让你求出x,y两数中第k大的公约数
思路：比最大公约数小的公约数都是它的因子，所以求出最大公约数的因子从大到小的第k个。
AC代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
#define ll long long
ll x,y,k;
const int N = 1e5;
ll s[N];
ll gcd(ll a, ll b)
{
    return a%b == 0 ? b : gcd(b, a%b);
}
int main()
{
    int t;
    scanf("%d",&t);

    while(t--)
    {
        memset(s,0,sizeof(s));
        scanf("%lld%lld%lld",&x,&y,&k);
        ll p=gcd(x,y);
        ll n=0;
        for(ll i=1;i*i<=p;i++)
        {
            if(p%i==0)
            {
               s[n++]=i;
               if(i*i!=p)
                s[n++]=p/i;
            }

        }
        if(n<k)
        {
            printf("-1\n");
            continue;
        }
        sort(s,s+n);
            printf("%lld\n",s[n-k]);


    }
}