Hdu5019revenge of gcd (enumeration + gcd)

Question link:

HuangJing

Question:
Find the GCD with the K value of two numbers
Ideas:
First, find the maximum common divisor. My first thought was to create a large prime number table, and then continue division to find the largest K number, but keep re, later, we know that we can directly calculate all the dikes of the maximum common divisor and sort them .. Then, the vector in STL is a good choice because the approximate number is unknown. Idea: Revenge of GCD Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 933 accepted submission (s): 282


Problem descriptionin 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.
Inputthe 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
 
Outputfor each test case, output the k-th GCD of X and Y. If no such integer exists, output-1.
Sample Input

32 3 12 3 28 16 3

 
Sample output

1-12

 
Sourcebestcoder round #10
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Code:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<vector>#include<cmath>#include<string>#include<queue>#define eps 1e-9#define ll long long#define INF 0x3f3f3f3fusing namespace std;ll xx,yy,kk;ll gcd(ll a,ll b){    if(b==0)  return a;    else  return gcd(b,a%b);}vector<ll>vec;int main(){     ll i;     int t;     scanf("%d",&t);     while(t--)     {        scanf("%I64d%I64d%I64d",&xx,&yy,&kk);        ll  ans=gcd(xx,yy);        vec.clear();        vec.push_back(1);        if(ans!=1)            vec.push_back(ans);        if(i*i==ans)            vec.push_back(i);        for(i=2;i*i<ans;i++)          {              if(ans%i==0)              {                  vec.push_back(i);                  vec.push_back((ll)ans/i);              }          }        sort(vec.begin(),vec.end());       // printf("%d\n",vec.size());        if(kk>vec.size())              printf("-1\n");        else              printf("%I64d\n",vec[vec.size()-kk]);    }     return 0;}

Hdu5019revenge of gcd (enumeration + gcd)