给定范围内偶数因子的元素数量
给定一个范围[n，m]，任务是找出在给定范围(包括 n 和 m)内具有偶数个因子的元素的数量。 例:
Input: n = 5, m = 20
Output: 14
The numbers with even factors are
5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20.
Input: n = 5, m = 100
Output: 88
一个简单的解决方法是从 n 开始循环所有的数字，对于每个数字，检查它是否有偶数个因子。如果它有偶数个因子，那么增加这些数字的计数，最后打印这些元素的数量。要高效地找到一个自然数的所有除数，请参考自然数的所有除数 一个高效的解决方案是找到具有奇数因子的数，即只有完美正方形具有奇数因子，因此除了完美正方形之外的所有数字都将具有偶数因子。所以，找出范围内完美方块的个数，从总数中减去，即 m-n+1 。 以下是上述方法的实施:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the perfect squares
int countOddSquares(int n, int m)
{
return (int)pow(m, 0.5) - (int)pow(n - 1, 0.5);
}
// Driver code
int main()
{
int n = 5, m = 100;
cout << "Count is "
<< (m - n + 1) - countOddSquares(n, m);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.io.*;
class GFG
{
// Function to count the perfect squares
static int countOddSquares(int n, int m)
{
return (int)Math.pow(m, 0.5) -
(int)Math.pow(n - 1, 0.5);
}
// Driver code
public static void main (String[] args)
{
int n = 5, m = 100;
System.out.println("Count is " + ((m - n + 1)
- countOddSquares(n, m)));
}
}
// This code is contributed by ajit..
Python 3
# Python3 implementation of the
# above approach
# Function to count the perfect squares
def countOddSquares(n, m) :
return (int(pow(m, 0.5)) -
int(pow(n - 1, 0.5)))
# Driver code
if __name__ == "__main__" :
n = 5 ; m = 100;
print("Count is", (m - n + 1) -
countOddSquares(n, m))
# This code is contributed by Ryuga
C
// C# implementation of the above approach
using System;
class GFG
{
// Function to count the perfect squares
static int countOddSquares(int n, int m)
{
return (int)Math.Pow(m, 0.5) -
(int)Math.Pow(n - 1, 0.5);
}
// Driver code
static public void Main ()
{
int n = 5, m = 100;
Console.WriteLine("Count is " + ((m - n + 1)
- countOddSquares(n, m)));
}
}
// This Code is contributed by akt_mit.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the
// above approach
// Function to count the perfect squares
function countOddSquares($n, $m)
{
return (int)pow($m, 0.5) -
(int)pow($n - 1, 0.5);
}
// Driver code
$n = 5;
$m = 100;
echo "Count is ", ($m - $n + 1) -
countOddSquares($n, $m);
// This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript implementation of the above approach
// Function to count the perfect squares
function countOddSquares(n, m)
{
return Math.pow(m,0.5) - Math.pow(n - 1, 0.5);
}
// Driver code
var n = 5, m = 100;
document.write( "Count is "
+ ((m - n + 1) - countOddSquares(n, m)));
</script>
Output:
Count is 88
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