hdu 5019(第K大公约数)
Posted
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了hdu 5019(第K大公约数)相关的知识,希望对你有一定的参考价值。
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
Today, GCD takes revenge on you. You have to figure out the k-th GCD of X and Y.
---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
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
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==0?a:gcd(b,a%b); } LL p[10005]; 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 = 0; for(LL i=1;i*i<=d;i++){ ///筛选出所有因子 if(d%i==0){ 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-1]); } } return 0; }
以上是关于hdu 5019(第K大公约数)的主要内容,如果未能解决你的问题,请参考以下文章