最大值(A，B)的最小可能值，使得 LCM(A，B) = C
给定一个整数 C ，任务是找到 max(A，B) 的最小可能值，使得 LCM(A，B) = C 。 举例:
输入: C = 6 输出: 3 最大值(1，6) = 6 最大值(2，3) = 3 和最小值(6，3) = 3 输入: C = 9 输出: 9
方法:解决这个问题的一个方法是使用这篇文章中讨论的方法找到给定数的所有因子，然后找到满足给定条件的对 (A，B) ，并取这些对的最大值的整体最小值。 以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the LCM of a and b
int lcm(int a, int b)
{
return (a / __gcd(a, b) * b);
}
// Function to return the minimized value
int getMinValue(int c)
{
int ans = INT_MAX;
// To find the factors
for (int i = 1; i <= sqrt(c); i++) {
// To check if i is a factor of c and
// the minimum possible number
// satisfying the given conditions
if (c % i == 0 && lcm(i, c / i) == c) {
// Update the answer
ans = min(ans, max(i, c / i));
}
}
return ans;
}
// Driver code
int main()
{
int c = 6;
cout << getMinValue(c);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class Solution
{
// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// Function to return the LCM of a and b
static int lcm(int a, int b)
{
return (a / __gcd(a, b) * b);
}
// Function to return the minimized value
static int getMinValue(int c)
{
int ans = Integer.MAX_VALUE;
// To find the factors
for (int i = 1; i <= Math.sqrt(c); i++)
{
// To check if i is a factor of c and
// the minimum possible number
// satisfying the given conditions
if (c % i == 0 && lcm(i, c / i) == c)
{
// Update the answer
ans = Math.min(ans, Math.max(i, c / i));
}
}
return ans;
}
// Driver code
public static void main(String args[])
{
int c = 6;
System.out.println(getMinValue(c));
}
}
// This code is contributed by Arnab Kundu
Python 3
# Python implementation of the approach
import sys
# Recursive function to return gcd of a and b
def __gcd(a, b):
# Everything divides 0
if (a == 0):
return b;
if (b == 0):
return a;
# base case
if (a == b):
return a;
# a is greater
if (a > b):
return __gcd(a - b, b);
return __gcd(a, b - a);
# Function to return the LCM of a and b
def lcm(a, b):
return (a / __gcd(a, b) * b);
# Function to return the minimized value
def getMinValue(c):
ans = sys.maxsize;
# To find the factors
for i in range(1, int(pow(c, 1/2)) + 1):
# To check if i is a factor of c and
# the minimum possible number
# satisfying the given conditions
if (c % i == 0 and lcm(i, c / i) == c):
# Update the answer
ans = min(ans, max(i, c / i));
return int(ans);
# Driver code
if __name__ == '__main__':
c = 6;
print(getMinValue(c));
# This code is contributed by 29AjayKumar
C
// C# implementation of the approach
using System;
class GFG
{
// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// Function to return the LCM of a and b
static int lcm(int a, int b)
{
return (a / __gcd(a, b) * b);
}
// Function to return the minimized value
static int getMinValue(int c)
{
int ans = int.MaxValue;
// To find the factors
for (int i = 1; i <= Math.Sqrt(c); i++)
{
// To check if i is a factor of c and
// the minimum possible number
// satisfying the given conditions
if (c % i == 0 && lcm(i, c / i) == c)
{
// Update the answer
ans = Math.Min(ans, Math.Max(i, c / i));
}
}
return ans;
}
// Driver code
public static void Main()
{
int c = 6;
Console.WriteLine(getMinValue(c));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// javascript implementation of the approach
// Recursive function to return gcd of a and b
function __gcd(a , b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// Function to return the LCM of a and b
function lcm(a , b)
{
return (a / __gcd(a, b) * b);
}
// Function to return the minimized value
function getMinValue(c)
{
var ans = Number.MAX_VALUE;
// To find the factors
for (i = 1; i <= Math.sqrt(c); i++)
{
// To check if i is a factor of c and
// the minimum possible number
// satisfying the given conditions
if (c % i == 0 && lcm(i, c / i) == c)
{
// Update the answer
ans = Math.min(ans, Math.max(i, c / i));
}
}
return ans;
}
// Driver code
var c = 6;
document.write(getMinValue(c));
// This code contributed by shikhasingrajput
</script>
Output:
3
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