找到第 n 个马赛克号
原文:https://www.geeksforgeeks.org/find-the-nth-mosaic-number/
给定一个整数 N ，任务是找到NthT5马赛克号。一个马赛克号可以表示如下: 如果N = AA BB CC…其中 A ， B ， C ..是 N 的质因数，那么第 N 个马赛克号将是A * A * B * B * C * C……。
示例:
输入: N = 8 输出: 6 8 可以表示为 2 3 。 所以，第 8 个马赛克号将是 2 * 3 = 6
输入: N = 36 输出: 24 36 可表示为 2 2 * 3 2 。 2 * 2 * 3 * 2 = 24
方法:我们必须通过将数除以因子，直到因子除以数，来找到数中所有的素因子以及因子的幂。第号号号镶嵌号将是已发现的主要因素及其力量的产物。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the nth mosaic number
int mosaic(int n)
{
int i, ans = 1;
// Iterate from 2 to the number
for (i = 2; i <= n; i++) {
// If i is the factor of n
if (n % i == 0 && n > 0) {
int count = 0;
// Find the count where i^count
// is a factor of n
while (n % i == 0) {
// Divide the number by i
n /= i;
// Increase the count
count++;
}
// Multiply the answer with
// count and i
ans *= count * i;
}
}
// Return the answer
return ans;
}
// Driver code
int main()
{
int n = 36;
cout << mosaic(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function to return the nth mosaic number
static int mosaic(int n)
{
int i, ans = 1;
// Iterate from 2 to the number
for (i = 2; i <= n; i++)
{
// If i is the factor of n
if (n % i == 0 && n > 0)
{
int count = 0;
// Find the count where i^count
// is a factor of n
while (n % i == 0)
{
// Divide the number by i
n /= i;
// Increase the count
count++;
}
// Multiply the answer with
// count and i
ans *= count * i;
}
}
// Return the answer
return ans;
}
// Driver code
public static void main (String[] args)
{
int n = 36;
System.out.println (mosaic(n));
}
}
// This code is contributed by jit_t.
Python 3
# Python3 implementation of the approach
# Function to return the nth mosaic number
def mosaic(n):
i=0
ans = 1
# Iterate from 2 to the number
for i in range(2,n+1):
# If i is the factor of n
if (n % i == 0 and n > 0):
count = 0
# Find the count where i^count
# is a factor of n
while (n % i == 0):
# Divide the number by i
n //= i
# Increase the count
count+=1
# Multiply the answer with
# count and i
ans *= count * i
# Return the answer
return ans
# Driver code
n = 36
print(mosaic(n))
# This code is contributed by mohit kumar 29
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the nth mosaic number
static int mosaic(int n)
{
int i, ans = 1;
// Iterate from 2 to the number
for (i = 2; i <= n; i++)
{
// If i is the factor of n
if (n % i == 0 && n > 0)
{
int count = 0;
// Find the count where i^count
// is a factor of n
while (n % i == 0)
{
// Divide the number by i
n /= i;
// Increase the count
count++;
}
// Multiply the answer with
// count and i
ans *= count * i;
}
}
// Return the answer
return ans;
}
// Driver code
static public void Main ()
{
int n = 36;
Console.WriteLine(mosaic(n));
}
}
// This code is contributed by ajit..
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the nth mosaic number
function mosaic(n)
{
let i, ans = 1;
// Iterate from 2 to the number
for(i = 2; i <= n; i++)
{
// If i is the factor of n
if (n % i == 0 && n > 0)
{
let count = 0;
// Find the count where i^count
// is a factor of n
while (n % i == 0)
{
// Divide the number by i
n = parseInt(n / i, 10);
// Increase the count
count++;
}
// Multiply the answer with
// count and i
ans *= count * i;
}
}
// Return the answer
return ans;
}
// Driver code
let n = 36;
document.write(mosaic(n));
// This code is contributed by mukesh07
</script>
Output:
24
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