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hdu 5019 Revenge of GCD（BestCoder Round #10)

2024-07-23 12:38:26   219人阅读

Revenge of GCD

                                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                    Total Submission(s): 668    Accepted Submission(s): 195


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
 

求第k大公约数。
求出最大公约数，枚举它的所有因子就行了。

代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
vector<long long> s;
long long gcd(long long a,long long b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    long long x,y,k;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        s.clear();
        scanf("%I64d%I64d%I64d",&x,&y,&k);
        long long u=gcd(x,y);
        for(long long i=1;i*i<=u;i++)
        {
            if(u%i==0)
            {
              s.push_back(i);
              if(i*i!=u)
              s.push_back(u/i);
            }
        }
        if(s.size()<k)
        {
            printf("-1\n");
        }
        else
        {
            sort(s.begin(),s.end());
            printf("%I64d\n",s[s.size()-k]);
        }
    }
    return 0;
}



hdu 5019 Revenge of GCD（BestCoder Round #10)


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