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HDU 5019 Revenge of GCD(数学)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5019
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
BestCoder Round #10
官方题解:
代码如下:
#include <cstdio> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <vector> using namespace std; typedef __int64 LL; vector<LL>v; LL GCD(LL a, LL b) { if(b == 0) return a; return GCD(b,a%b); } int main() { int t; LL x, y, k; scanf("%d",&t); while(t--) { v.clear(); scanf("%I64d%I64d%I64d",&x,&y,&k); LL tt = GCD(x,y); // printf("%I64d\n",tt); for(LL i = 1; i*i <= tt; i++) { if(tt%i == 0) { v.push_back(i); if(i*i != tt)//防止放入两个i v.push_back(tt/i); } } sort(v.begin(), v.end()); if(k > v.size()) printf("-1\n"); else printf("%I64d\n",v[v.size()-k]); } return 0; }
HDU 5019 Revenge of GCD(数学)
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