求[1，n]
范围内所有数的除数
给定一个整数 N 。任务是找出【1，N】范围内所有数字的除数。
示例:
输入: N = 5 输出: 1 2 2 3 2 除数(1) = 1 除数(2) = 1 和 2 除数(3) = 1 和 3 除数(4) = 1、2 和 4 除数(5) = 1 和 5
输入: N = 10 输出:1 2 2 3 4 2 4 3 4
方法:创建一个大小为 (N + 1) 的数组 arr[] ，其中 arr[i] 存储 i 的除数。现在，对于范围【1，N】中的每一个 j ，递增所有可被 j 整除的元素。 例如，如果 j = 3 则更新 arr[3]、arr[6]、arr[9]，…
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 find the number of divisors
// of all numbers in the range [1, n]
void findDivisors(int n)
{
// Array to store the count
// of divisors
int div[n + 1];
memset(div, 0, sizeof div);
// For every number from 1 to n
for (int i = 1; i <= n; i++) {
// Increase divisors count for
// every number divisible by i
for (int j = 1; j * i <= n; j++)
div[i * j]++;
}
// Print the divisors
for (int i = 1; i <= n; i++)
cout << div[i] << " ";
}
// Driver code
int main()
{
int n = 10;
findDivisors(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
class GFG
{
// Function to find the number of divisors
// of all numbers in the range [1, n]
static void findDivisors(int n)
{
// Array to store the count
// of divisors
int[] div = new int[n + 1];
// For every number from 1 to n
for (int i = 1; i <= n; i++)
{
// Increase divisors count for
// every number divisible by i
for (int j = 1; j * i <= n; j++)
div[i * j]++;
}
// Print the divisors
for (int i = 1; i <= n; i++)
System.out.print(div[i]+" ");
}
// Driver code
public static void main(String args[])
{
int n = 10;
findDivisors(n);
}
}
// This code is contributed by Ryuga
Python 3
# Python3 implementation of the approach
# Function to find the number of divisors
# of all numbers in the range [1,n]
def findDivisors(n):
# List to store the count
# of divisors
div = [0 for i in range(n + 1)]
# For every number from 1 to n
for i in range(1, n + 1):
# Increase divisors count for
# every number divisible by i
for j in range(1, n + 1):
if j * i <= n:
div[i * j] += 1
# Print the divisors
for i in range(1, n + 1):
print(div[i], end = " ")
# Driver Code
if __name__ == "__main__":
n = 10
findDivisors(n)
# This code is contributed by
# Vivek Kumar Singh
C
// C# implementation of the approach
using System;
class GFG
{
// Function to find the number of divisors
// of all numbers in the range [1, n]
static void findDivisors(int n)
{
// Array to store the count
// of divisors
int[] div = new int[n + 1];
// For every number from 1 to n
for (int i = 1; i <= n; i++)
{
// Increase divisors count for
// every number divisible by i
for (int j = 1; j * i <= n; j++)
div[i * j]++;
}
// Print the divisors
for (int i = 1; i <= n; i++)
Console.Write(div[i]+" ");
}
// Driver code
static void Main()
{
int n = 10;
findDivisors(n);
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to find the number of divisors
// of all numbers in the range [1, n]
function findDivisors($n)
{
// Array to store the count
// of divisors
$div = array_fill(0, $n + 2, 0);
// For every number from 1 to n
for ($i = 1; $i <= $n; $i++)
{
// Increase divisors count for
// every number divisible by i
for ($j = 1; $j * $i <= $n; $j++)
$div[$i * $j]++;
}
// Print the divisors
for ($i = 1; $i <= $n; $i++)
echo $div[$i], " ";
}
// Driver code
$n = 10;
findDivisors($n);
// This code is contributed
// by Arnab Kundu
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to find the number of divisors
// of all numbers in the range [1, n]
function findDivisors(n)
{
// Array to store the count
// of divisors
let div = new Array(n + 1).fill(0);
// For every number from 1 to n
for(let i = 1; i <= n; i++)
{
// Increase divisors count for
// every number divisible by i
for(let j = 1; j * i <= n; j++)
div[i * j]++;
}
// Print the divisors
for(let i = 1; i <= n; i++)
document.write(div[i] + " ");
}
// Driver code
let n = 10;
findDivisors(n);
// This code is contributed by souravmahato348
</script>
Output:
1 2 2 3 2 4 2 4 3 4
版权属于：月萌API www.moonapi.com，转载请注明出处
