GDUT_寒假训练题解报告_数论专题_个人题解报告——题目：E - Revenge of GCD【HDU - 5019】【数论（GCD，因数分解）】

最新推荐文章于 2020-04-06 03:28:25 发布

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本文介绍了一种求解两个整数X和Y的第K个最大公因数(GCD)的算法。通过计算X和Y的最大公因数，找到其所有因数并排序，从而确定第K个GCD。适用于ACM竞赛等场景。

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=5019
题目：

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

这个玩意，我因为long long又死了一遍，其实不难，就是最大公因数的第k个因数就行了。

虽然是这么说，但是没有卡极限数据感觉，如果极限数据，我找所有因数的时候，用的是暴力，time*time<=GCD(a,b)，也就是次数time最大可以达到1e6，然后t范围是1e2，最大是1e8已经要爆表了

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>
#include <queue>
#include <stack>
#include <map>
//鬼畜头文件
using namespace std;
#define INF 0x3f3f3f3f
#define LLU unsigned long long
#define LL long long
const int mod =1e9+7;
//鬼畜define
int t;
LL a,b,k;

LL gcd(LL a,LL b)
{
	if(b==0)return a;
	return gcd(b,a%b);
}
bool cmp(LL a,LL b)
{
	return a>b;
}
LL all[10000];
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		int ans=1;
		scanf("%lld %lld %lld",&a,&b,&k);
		LL point=gcd(a,b);
		int num=0;
		LL time;
		for(time=1;time*time<point;time++)
		{
			if(point%time==0)
			{
				all[num++]=time;
				all[num++]=point/time;
			}
		}
		if(time*time==point)all[num++]=time;

		if(num>=k){sort(all,all+num,cmp);printf("%lld\n",all[k-1]);}
		else printf("-1\n");
	}


	return 0;
}