hdu 5019(第K大公约数)
2024-09-08 16:13:31
Revenge of GCD
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2140 Accepted Submission(s): 596
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
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
2 3 1
2 3 2
8 16 3
Sample Output
1
-1
2
-1
2
Source
被坑惨了。。GCD的参数传的INT。。求出最大公约数然后求最大公约数的第K大因子就好。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iostream>
using namespace std;
typedef long long LL;
LL gcd(LL a,LL b){
return b==?a:gcd(b,a%b);
}
LL p[];
LL cmp(LL a,LL b){
return a>b;
}
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--){
LL a,b,k;
scanf("%lld%lld%lld",&a,&b,&k);
LL d = gcd(a,b);
int id = ;
for(LL i=;i*i<=d;i++){ ///筛选出所有因子
if(d%i==){
if(i*i==d) p[id++]=i;
else{
p[id++]=i;
p[id++]=d/i;
}
}
}
if(id<k){
printf("-1\n");
}
else{
sort(p,p+id,cmp);
printf("%lld\n",p[k-]);
}
}
return ;
}
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