Codeforces--1285C--Fadi and LCM

题目描述：
Today, Osama gave Fadi an integer X, and Fadi was wondering about the minimum possible value of max(a,b) such that LCM(a,b) equals X. Both a and b should be positive integers.
LCM(a,b) is the smallest positive integer that is divisible by both a and b. For example, LCM(6,8)=24, LCM(4,12)=12, LCM(2,3)=6.
Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?
输入描述:
The first and only line contains an integer X (1≤X≤1012).
输出描述:
Print two positive integers, a and b, such that the value of max(a,b) is minimum possible and LCM(a,b) equals X. If there are several possible such pairs, you can print any.
输入：
2
6
4
1
输出：
1 2
2 3
1 4
1 1
题意：
请你找到这样的一对数(a,b),使得LCM(a,b)=x,并且max(a,b)要尽可能的小
题解：
两个数尽可能接近且两个数互质，这样max(a,b)才会越小，那就从sqrt(x)开始枚举，找到符合的输出即可
代码：

#include
#include
#include
#include
#include
using namespace std;typedef long long ll;ll gcd(ll a,ll b){return b ? gcd(b,a % b) : a;
}int main(){ll x;while(scanf("%lld",&x)!=EOF){ll ans1 = 0,ans2 = 0;ll k = (ll)sqrt(x + 0.5);for(ll i = k; i >= 1; i --){if(x % i == 0){ans1 = i;ans2 = x / i;if(gcd(ans1,ans2) == 1) break;}}printf("%lld %lld\n",min(ans1,ans2),max(ans1,ans2));}return 0;
}

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