求一个数的偶数因子之和
原文:https://www.geeksforgeeks.org/find-sum-even-factors-number/
给定一个数 n，任务是求一个数的偶因子和。 示例:
Input : 30
Output : 48
Even dividers sum 2 + 6 + 10 + 30 = 48
Input : 18
Output : 26
Even dividers sum 2 + 6 + 18 = 26
先决条件:因子之和 如前所述，一个数的因子之和为 设 p1，p2，… pk 为 n 的素因子，设 a1，a2，..ak 是 p1，p2，..pk 分别除以 n，即我们可以把 n 写成n =(P1a1)(p2a2)……(PKAK)。
Sum of divisors = (1 + p1 + p12 ... p1a1) *
(1 + p2 + p22 ... p2a2) *
...........................
(1 + pk + pk2 ... pkak)
如果数是奇数，那么没有偶数因子，所以我们简单地返回 0。 如果数是偶数，我们用上面的公式。我们只需要忽略 2 0 。所有其他项相乘产生偶数因子和。例如，考虑 n = 18。可以写成 2 1 3 2 全因子的太阳是(20+21)(30+31+32)。如果我们去掉 2 0 ，那么我们得到的是 偶数因子之和(2)(1+3+3 2 ) = 26。 去掉偶数因子中的奇数，我们忽略那么 2 0 whaich 就是 1。经过这一步，我们只得到偶数因子。注意 2 是唯一的偶数素数。 以下是上述办法的实施情况。
C++
// Formula based CPP program to find sum of all
// divisors of n.
#include <bits/stdc++.h>
using namespace std;
// Returns sum of all factors of n.
int sumofFactors(int n)
{
// If n is odd, then there are no even factors.
if (n % 2 != 0)
return 0;
// Traversing through all prime factors.
int res = 1;
for (int i = 2; i <= sqrt(n); i++) {
// While i divides n, print i and divide n
int count = 0, curr_sum = 1, curr_term = 1;
while (n % i == 0) {
count++;
n = n / i;
// here we remove the 2^0 that is 1. All
// other factors
if (i == 2 && count == 1)
curr_sum = 0;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
// This condition is to handle the case when n
// is a prime number.
if (n >= 2)
res *= (1 + n);
return res;
}
// Driver code
int main()
{
int n = 18;
cout << sumofFactors(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Formula based Java program to
// find sum of all divisors of n.
import java.util.*;
import java.lang.*;
public class GfG{
// Returns sum of all factors of n.
public static int sumofFactors(int n)
{
// If n is odd, then there
// are no even factors.
if (n % 2 != 0)
return 0;
// Traversing through all prime
// factors.
int res = 1;
for (int i = 2; i <= Math.sqrt(n); i++)
{
int count = 0, curr_sum = 1;
int curr_term = 1;
// While i divides n, print i and
// divide n
while (n % i == 0)
{
count++;
n = n / i;
// here we remove the 2^0 that
// is 1\. All other factors
if (i == 2 && count == 1)
curr_sum = 0;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
// This condition is to handle the
// case when n is a prime number.
if (n >= 2)
res *= (1 + n);
return res;
}
// Driver function
public static void main(String argc[]){
int n = 18;
System.out.println(sumofFactors(n));
}
}
/* This code is contributed by Sagar Shukla */
Python 3
# Formula based Python3
# program to find sum
# of alldivisors of n.
import math
# Returns sum of all
# factors of n.
def sumofFactors(n) :
# If n is odd, then
# there are no even
# factors.
if (n % 2 != 0) :
return 0
# Traversing through
# all prime factors.
res = 1
for i in range(2, (int)(math.sqrt(n)) + 1) :
# While i divides n
# print i and divide n
count = 0
curr_sum = 1
curr_term = 1
while (n % i == 0) :
count= count + 1
n = n // i
# here we remove the
# 2^0 that is 1\. All
# other factors
if (i == 2 and count == 1) :
curr_sum = 0
curr_term = curr_term * i
curr_sum = curr_sum + curr_term
res = res * curr_sum
# This condition is to
# handle the case when
# n is a prime number.
if (n >= 2) :
res = res * (1 + n)
return res
# Driver code
n = 18
print(sumofFactors(n))
# This code is contributed by Nikita Tiwari.
C
// Formula based C# program to
// find sum of all divisors of n.
using System;
public class GfG {
// Returns sum of all factors of n.
public static int sumofFactors(int n)
{
// If n is odd, then there
// are no even factors.
if (n % 2 != 0)
return 0;
// Traversing through all prime factors.
int res = 1;
for (int i = 2; i <= Math.Sqrt(n); i++)
{
int count = 0, curr_sum = 1;
int curr_term = 1;
// While i divides n, print i
// and divide n
while (n % i == 0)
{
count++;
n = n / i;
// here we remove the 2^0 that
// is 1\. All other factors
if (i == 2 && count == 1)
curr_sum = 0;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
// This condition is to handle the
// case when n is a prime number.
if (n >= 2)
res *= (1 + n);
return res;
}
// Driver Code
public static void Main() {
int n = 18;
Console.WriteLine(sumofFactors(n));
}
}
// This code is contributed by vt_m
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Formula based php program to find sum
// of all divisors of n.
// Returns sum of all factors of n.
function sumofFactors($n)
{
// If n is odd, then there are no
// even factors.
if ($n % 2 != 0)
return 0;
// Traversing through all prime factors.
$res = 1;
for ($i = 2; $i <= sqrt($n); $i++) {
// While i divides n, print i
// and divide n
$count = 0;
$curr_sum = 1;
$curr_term = 1;
while ($n % $i == 0) {
$count++;
$n = floor($n / $i);
// here we remove the 2^0
// that is 1\. All other
// factors
if ($i == 2 && $count == 1)
$curr_sum = 0;
$curr_term *= $i;
$curr_sum += $curr_term;
}
$res *= $curr_sum;
}
// This condition is to handle the
// case when n is a prime number.
if ($n >= 2)
$res *= (1 + $n);
return $res;
}
// Driver code
$n = 18;
echo sumofFactors($n);
// This code is contributed by mits
?>
java 描述语言
<script>
// javascript program to
// find sum of all divisors of n.
// Returns sum of all factors of n.
function sumofFactors(n)
{
// If n is odd, then there
// are no even factors.
if (n % 2 != 0)
return 0;
// Traversing through all prime
// factors.
let res = 1;
for (let i = 2; i <= Math.sqrt(n); i++)
{
let count = 0, curr_sum = 1;
let curr_term = 1;
// While i divides n, print i and
// divide n
while (n % i == 0)
{
count++;
n = n / i;
// here we remove the 2^0 that
// is 1\. All other factors
if (i == 2 && count == 1)
curr_sum = 0;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
// This condition is to handle the
// case when n is a prime number.
if (n >= 2)
res *= (1 + n);
return res;
}
// Driver Function
let n = 18;
document.write(sumofFactors(n));
// This code is contributed by susmitakundugoaldanga.
</script>
输出:
26
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