求最小值 x，使得(x % k)*(x/k)= n | Set-2
原文:https://www . geesforgeks . org/find-minimum-x-so-x-k-x-k-n-set-2/
给定两个正整数 n 和 k，求最小正整数 x，使得(x % k) * (x / k) == n，其中%是模运算符，/是整数除法运算符。 例:
输入: n = 4，k = 6 输出: 10 解释: (10 % 6) * (10 / 6) = (4) * (1) = 4 等于 n 输入: n = 5，k = 5 输出: 26
方法:使用集合-1 中的 K 值解决了问题。在本文中，我们将使用因子方法来解决上述问题。下面给出了解决上述问题的步骤。
- 因为方程的形式是 a*b = N，所以 a 和 b 是 N 的因子。
- 从 1 迭代到 sqrt(N)得到所有的因子。
- 因子 I 和 n/i 可以是 A，也可以是 b。
- 如果 I 是 A，n/i 是 B，那么这个数就是 i*k + (n/i) 。我们可以通过与 N 比较来检查这个数是否是 1，如果是，那么它就是一个满足方程的数。
- 如果 I 是 B，n/i 是 A，那么这个数字就是 (n/i)*k + (i) 。我们可以通过与 N 比较来检查这个数是否是 1，如果是，那么它就是一个满足方程的数。
- 获取所有满足方程的数字，并打印其中最小的数字。
下面是上述方法的实现。
C++
// CPP Program to find the minimum
// positive X such that the given
// equation holds true
#include <bits/stdc++.h>
using namespace std;
// This function gives the required
// answer
int minimumX(int n, int k)
{
int mini = INT_MAX;
// Iterate for all the factors
for (int i = 1; i * i <= n; i++) {
// Check if i is a factor
if (n % i == 0) {
int fir = i;
int sec = n / i;
int num1 = fir * k + sec;
// Consider i to be A and n/i to be B
int res = (num1 / k) * (num1 % k);
if (res == n)
mini = min(num1, mini);
int num2 = sec * k + fir;
res = (num2 / k) * (num2 % k);
// Consider i to be B and n/i to be A
if (res == n)
mini = min(num2, mini);
}
}
return mini;
}
// Driver Code to test above function
int main()
{
int n = 4, k = 6;
cout << minimumX(n, k) << endl;
n = 5, k = 5;
cout << minimumX(n, k) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to find the minimum
// positive X such that the given
// equation holds true
import java.util.*;
class solution
{
// This function gives the required
// answer
static int minimumX(int n, int k)
{
int mini = Integer.MAX_VALUE;
// Iterate for all the factors
for (int i = 1; i * i <= n; i++) {
// Check if i is a factor
if (n % i == 0) {
int fir = i;
int sec = n / i;
int num1 = fir * k + sec;
// Consider i to be A and n/i to be B
int res = (num1 / k) * (num1 % k);
if (res == n)
mini = Math.min(num1, mini);
int num2 = sec * k + fir;
res = (num2 / k) * (num2 % k);
// Consider i to be B and n/i to be A
if (res == n)
mini = Math.min(num2, mini);
}
}
return mini;
}
// Driver Code to test above function
public static void main(String args[])
{
int n = 4, k = 6;
System.out.println(minimumX(n, k));
n = 5;
k = 5;
System.out.println(minimumX(n, k));
}
}
Python 3
# Python3 program to find the minimum
# positive X such that the given
# equation holds true
import sys
# This function gives the required
# answer
def minimumX(n, k):
mini = sys.maxsize
# Iterate for all the factors
i = 1
while i * i <= n :
# Check if i is a factor
if (n % i == 0) :
fir = i
sec = n // i
num1 = fir * k + sec
# Consider i to be A and n/i to be B
res = (num1 // k) * (num1 % k)
if (res == n):
mini = min(num1, mini)
num2 = sec * k + fir
res = (num2 // k) * (num2 % k)
# Consider i to be B and n/i to be A
if (res == n):
mini = min(num2, mini)
i += 1
return mini
# Driver Code
if __name__ == "__main__":
n = 4
k = 6
print (minimumX(n, k))
n = 5
k = 5
print (minimumX(n, k))
# This code is contributed by ita_c
C
// C# Program to find the minimum
// positive X such that the given
// equation holds true
using System;
class solution
{
// This function gives the required
// answer
static int minimumX(int n, int k)
{
int mini = int.MaxValue;
// Iterate for all the factors
for (int i = 1; i * i <= n; i++) {
// Check if i is a factor
if (n % i == 0) {
int fir = i;
int sec = n / i;
int num1 = fir * k + sec;
// Consider i to be A and n/i to be B
int res = (num1 / k) * (num1 % k);
if (res == n)
mini = Math.Min(num1, mini);
int num2 = sec * k + fir;
res = (num2 / k) * (num2 % k);
// Consider i to be B and n/i to be A
if (res == n)
mini = Math.Min(num2, mini);
}
}
return mini;
}
// Driver Code to test above function
public static void Main()
{
int n = 4, k = 6;
Console.WriteLine(minimumX(n, k));
n = 5;
k = 5;
Console.WriteLine(minimumX(n, k));
}
// This code is contributed by Ryuga
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Program to find the minimum
// positive X such that the given
// equation holds true
// Function gives the required
// answer
function minimumX($n, $k)
{
$mini = PHP_INT_MAX;
// Iterate for all the factors
for ($i = 1; $i * $i <= $n; $i++)
{
// Check if i is a factor
if ($n % $i == 0)
{
$fir = $i;
$sec = (int)$n / $i;
$num1 = $fir * $k + $sec;
// Consider i to be A and n/i to be B
$res = (int)($num1 / $k) *
($num1 % $k);
if ($res == $n)
$mini = min($num1, $mini);
$num2 = $sec * $k + $fir;
$res = (int)($num2 / $k) *
($num2 % $k);
// Consider i to be B and n/i to be A
if ($res == $n)
$mini = min($num2, $mini);
}
}
return $mini;
}
// Driver Code
$n = 4;
$k = 6;
echo minimumX($n, $k), "\n";
$n = 5;
$k = 5;
echo minimumX($n, $k), "\n";
// This code is contributed
// by Sach_Code
?>
java 描述语言
<script>
// Javascript Program to find the minimum
// positive X such that the given
// equation holds true
// This function gives the required
// answer
function minimumX(n, k)
{
let mini = Number.MAX_VALUE;
// Iterate for all the factors
for (let i = 1; i * i <= n; i++) {
// Check if i is a factor
if (n % i == 0) {
let fir = i;
let sec = parseInt(n / i, 10);
let num1 = fir * k + sec;
// Consider i to be A and n/i to be B
let res = parseInt((num1 / k), 10) * (num1 % k);
if (res == n)
mini = Math.min(num1, mini);
let num2 = sec * k + fir;
res = parseInt((num2 / k), 10) * (num2 % k);
// Consider i to be B and n/i to be A
if (res == n)
mini = Math.min(num2, mini);
}
}
return mini;
}
let n = 4, k = 6;
document.write(minimumX(n, k) + "<br>");
n = 5, k = 5;
document.write(minimumX(n, k));
</script>
Output:
10
26
时间复杂度: O(sqrt(N))，其中 N 为给定正整数。
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