首页 > 代码库 > HDOJ 5019 Revenge of GCD

HDOJ 5019 Revenge of GCD

2024-07-23 17:18:06   217人阅读


第k大GCD = GCD／第K大因子

Revenge of GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 877    Accepted Submission(s): 259


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
 


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long int LL;

LL gcd(LL a,LL b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}

LL factor[1000100],n;

int main()
{
    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        LL x,y,k;
        cin>>x>>y>>k;
        LL g=gcd(x,y);
        LL d=sqrt(g);
        n=0;
        for(LL i=1;i<=d;i++)
        {
            if(g%i==0)
            {
                LL j=g/i;
                factor[n++]=i;
                if(j!=i) factor[n++]=j;
            }
        }
        sort(factor,factor+n);
        if(k>n) puts("-1");
        else
        {
            cout<<g/factor[k-1]<<endl;
        }
    }
    return 0;
}



HDOJ 5019 Revenge of GCD


联系
我们