求一个数与其最大质因数之和
给定一个整数 N ，任务是求 N 的和及其最大质因数。 例:
输入:19 T3】输出:38 19 的最大质因数为 19。 因此，19 + 19 = 38 输入: 8 输出: 10 8 + 2 = 10
逼近:找到该数的最大质因数存入 maxPrimeFact 然后打印 N + maxPrimeFact 的值。 以下是上述方法的实施:
C++
// C++ program to find sum of n and
// it's largest prime factor
#include <cmath>
#include <iostream>
using namespace std;
// Function to return the sum of n and
// it's largest prime factor
int maxPrimeFactors(int n)
{
int num = n;
// Initialise maxPrime to -1.
int maxPrime = -1;
while (n % 2 == 0) {
maxPrime = 2;
n /= 2;
}
// n must be odd at this point, thus skip
// the even numbers and iterate only odd numbers
for (int i = 3; i <= sqrt(n); i += 2) {
while (n % i == 0) {
maxPrime = i;
n = n / i;
}
}
// This condition is to handle the case
// when n is a prime number greater than 2
if (n > 2)
maxPrime = n;
// finally return the sum.
int sum = maxPrime + num;
return sum;
}
// Driver Program to check the above function.
int main()
{
int n = 19;
cout << maxPrimeFactors(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find sum of n and
// it's largest prime factor
import java.io.*;
class GFG
{
// Function to return the sum of n
// and it's largest prime factor
static int maxPrimeFactors(int n)
{
int num = n;
// Initialise maxPrime to -1.
int maxPrime = -1;
while (n % 2 == 0)
{
maxPrime = 2;
n /= 2;
}
// n must be odd at this point,
// thus skip the even numbers and
// iterate only odd numbers
for (int i = 3; i <= Math.sqrt(n); i += 2) {
while (n % i == 0) {
maxPrime = i; n = n / i;
}
}
// This condition is to handle the case
// when n is a prime number greater than 2
if (n > 2) {
maxPrime = n;
}
// finally return the sum.
int sum = maxPrime + num;
return sum;
}
// Driver Code
public static void main (String[] args)
{
int n = 19;
System.out.println(maxPrimeFactors(n));
}
}
// This code is contributed by anuj_67
Python 3
# Python 3 program to find sum of n and
# it's largest prime factor
from math import sqrt
# Function to return the sum of n and
# it's largest prime factor
def maxPrimeFactors(n):
num = n
# Initialise maxPrime to -1.
maxPrime = -1;
while (n % 2 == 0):
maxPrime = 2
n = n / 2
# n must be odd at this point, thus skip
# the even numbers and iterate only odd numbers
p = int(sqrt(n) + 1)
for i in range(3, p, 2):
while (n % i == 0):
maxPrime = i
n = n / i
# This condition is to handle the case
# when n is a prime number greater than 2
if (n > 2):
maxPrime = n
# finally return the sum.
sum = maxPrime + num
return sum
# Driver Code
if __name__ == '__main__':
n = 19
print(maxPrimeFactors(n))
# This code is contributed by
# Surendra_Gangwar
C
// C# program to find sum of n and
// it's largest prime factor
using System;
class GFG
{
// Function to return the sum of n
// and it's largest prime factor
static int maxPrimeFactors(int n)
{
int num = n;
// Initialise maxPrime to -1.
int maxPrime = -1;
while (n % 2 == 0)
{
maxPrime = 2;
n /= 2;
}
// n must be odd at this point,
// thus skip the even numbers and
// iterate only odd numbers
for (int i = 3;
i <= Math.Sqrt(n); i += 2)
{
while (n % i == 0)
{
maxPrime = i; n = n / i;
}
}
// This condition is to handle the case
// when n is a prime number greater than 2
if (n > 2)
{
maxPrime = n;
}
// finally return the sum.
int sum = maxPrime + num;
return sum;
}
// Driver Code
static void Main ()
{
int n = 19;
Console.WriteLine(maxPrimeFactors(n));
}
}
// 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 sum of n and
// it's largest prime factor
// Function to return the sum of n
// and it's largest prime factor
function maxPrimeFactors($n)
{
$num = $n;
// Initialise maxPrime to -1.
$maxPrime = -1;
while ($n % 2 == 0)
{
$maxPrime = 2;
$n /= 2;
}
// n must be odd at this point,
// thus skip the even numbers
// and iterate only odd numbers
for ($i = 3; $i <= sqrt($n); $i += 2)
{
while ($n % $i == 0)
{
$maxPrime = $i;
$n = $n / $i;
}
}
// This condition is to handle the case
// when n is a prime number greater than 2
if ($n > 2)
$maxPrime = $n;
// finally return the sum.
$sum = $maxPrime + $num;
return $sum;
}
// Driver Code
$n = 19;
echo maxPrimeFactors($n);
// This code is contributed
// by inder_verma
?>
java 描述语言
<script>
// Javascript program to find sum of n and
// it's largest prime factor
// Function to return the sum of n
// and it's largest prime factor
function maxPrimeFactors(n)
{
var num = n;
// Initialise maxPrime to -1.
var maxPrime = -1;
while (n % 2 == 0)
{
maxPrime = 2;
n = parseInt(n/2);
}
// n must be odd at this point,
// thus skip the even numbers and
// iterate only odd numbers
for (var i = 3; i <= parseInt(Math.sqrt(n)); i += 2)
{
while (n % i == 0) {
maxPrime = i;
n = parseInt(n / i);
}
}
// This condition is to handle the case
// when n is a prime number greater than 2
if (n > 2) {
maxPrime = n;
}
// finally return the sum.
var sum = maxPrime + num;
return sum;
}
// Driver Code
var n = 19;
document.write(maxPrimeFactors(n));
// This code contributed by shikhasingrajput
</script>
Output:
38
版权属于：月萌API www.moonapi.com，转载请注明出处
