给定一个数字,找出其因素的最小和。
例子:
Input : 12
Output : 7
Explanation:
Following are different ways to factorize 12 and
sum of factors in different ways.
12 = 12 * 1 = 12 + 1 = 13
12 = 2 * 6 = 2 + 6 = 8
12 = 3 * 4 = 3 + 4 = 7
12 = 2 * 2 * 3 = 2 + 2 + 3 = 7
Therefore minimum sum is 7
Input : 105
Output : 15
为了使总和最小化,我们必须尽可能长地分解因数。通过这个过程,我们成为主要因素。因此,为了找到数量乘积的最小和,我们找到乘积的素因之和。
C++
// CPP program to find minimum
// sum of product of number
#include
using namespace std;
// To find minimum sum of
// product of number
int findMinSum(int num)
{
int sum = 0;
// Find factors of number
// and add to the sum
for (int i = 2; i * i <= num; i++) {
while (num % i == 0) {
sum += i;
num /= i;
}
}
sum += num;
// Return sum of numbers
// having minimum product
return sum;
}
// Driver program to test above function
int main()
{
int num = 12;
cout << findMinSum(num);
return 0;
}
Java
// Java program to find minimum
// sum of product of number
public class Main {
// To find minimum sum of
// product of number
static int findMinSum(int num)
{
int sum = 0;
// Find factors of number
// and add to the sum
for (int i = 2; i * i <= num; i++) {
while (num % i == 0) {
sum += i;
num /= i;
}
}
sum += num;
// Return sum of numbers
// having minimum product
return sum;
}
// Driver program to test above function
public static void main(String[] args)
{
int num = 12;
System.out.println(findMinSum(num));
}
}
Python
# Python program to find minimum
# sum of product of number
# To find minimum sum of
# product of number
def findMinSum(num):
sum = 0
# Find factors of number
# and add to the sum
i = 2
while(i * i <= num):
while(num % i == 0):
sum += i
num /= i
i += 1
sum += num
# Return sum of numbers
# having minimum product
return sum
# Driver Code
num = 12
print findMinSum(num)
# This code is contributed by Sachin Bisht
C#
// C# program to find minimum
// sum of product of number
using System;
public class GFG {
// To find minimum sum of
// product of number
static int findMinSum(int num)
{
int sum = 0;
// Find factors of number
// and add to the sum
for (int i = 2; i * i <= num; i++) {
while (num % i == 0) {
sum += i;
num /= i;
}
}
sum += num;
// Return sum of numbers
// having minimum product
return sum;
}
// Driver Code
public static void Main()
{
int num = 12;
Console.Write(findMinSum(num));
}
}
// This Code is contributed by Nitin Mittal.
PHP
Javascript
输出:
7