从 1 到 N 选择包含偶数和奇数的配对的方式数
原文:https://www . geesforgeks . org/从 1 到 n 选择一对包含一个偶数和一个奇数的路数/
给定一个数字 N，任务是从 1 到 N(含 1 和 N)之间的数字中找出包含偶数和奇数的对的数量。 注:对中数字的顺序无关紧要。即(1，2)和(2，1)是相同的。
示例:
Input: N = 3
Output: 2
The pairs are (1, 2) and (2, 3).
Input: N = 6
Output: 9
The pairs are (1, 2), (1, 4), (1, 6), (2, 3),
(2, 5), (3, 4), (3, 6), (4, 5), (5, 6).
方式:成对的方式数为(偶数总数*奇数总数)。 如此
- 如果 N 是偶数=奇数= N/2
- 如果 N 是奇数，偶数= N/2，奇数= N/2+1
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include <iostream>
using namespace std;
// Driver code
int main()
{
int N = 6;
int Even = N / 2 ;
int Odd = N - Even ;
cout << Even * Odd ;
return 0;
// This code is contributed
// by ANKITRAI1
}
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.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// Driver code
public static void main(String args[])
{
int N = 6;
int Even = N / 2 ;
int Odd = N - Even ;
System.out.println( Even * Odd );
}
}
Python 3
# Python implementation of the above approach
N = 6
# number of even numbers
Even = N//2
# number of odd numbers
Odd = N-Even
print(Even * Odd)
C
// C# implementation of the
// above approach
using System;
class GFG
{
// Driver code
public static void Main()
{
int N = 6;
int Even = N / 2 ;
int Odd = N - Even ;
Console.WriteLine(Even * Odd);
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the
// above approach
// Driver code
$N = 6;
$Even = $N / 2 ;
$Odd = $N - $Even ;
echo $Even * $Odd ;
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// Javascript implementation of the above approach
// Driver code
let N = 6;
let Even = Math.floor(N / 2) ;
let Odd = N - Even ;
document.write( Even * Odd );
// This code is contributed by avanitrachhadiya2155
</script>
Output:
9
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