从 1 到 N 选择三个数字得到偶数和的方法数
原文:https://www . geesforgeks . org/number-way-get-even-sum-choice-三数-1-n/
给定一个整数 N，找出我们可以从{1，2，3 …，N}中选择 3 个数字的方法数，使它们的和为偶数。 示例:
Input : N = 3
Output : 1
Explanation: Select 1, 2 and 3
Input : N = 4
Output : 2
Either select (1, 2, 3) or (1, 3, 4)
为了得到和，只能有两种情况:
- 取 2 个奇数和 1 个偶数。
- 取所有偶数。
If n is even,
Count of odd numbers = n/2 and even = n/2.
Else
Count odd numbers = n/2 +1 and evne = n/2.
情况 1–路数为:奇数 C 2 偶数。 情况 2–路号为:偶数 C 3 。 所以，总的方式是 Case_1_result + Case_2_result。
C++
// C++ program for above implementation
#include <bits/stdc++.h>
#define MOD 1000000007
using namespace std;
// Function to count number of ways
int countWays(int N)
{
long long int count, odd = N / 2, even;
if (N & 1)
odd = N / 2 + 1;
even = N / 2;
// Case 1: 2 odds and 1 even
count = (((odd * (odd - 1)) / 2) * even) % MOD;
// Case 2: 3 evens
count = (count + ((even * (even - 1) *
(even - 2)) / 6)) % MOD;
return count;
}
// Driver code
int main()
{
int n = 10;
cout << countWays(n) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// java program for above implementation
import java.io.*;
class GFG {
static long MOD = 1000000007;
// Function to count number of ways
static long countWays(int N)
{
long count, odd = N / 2, even;
if ((N & 1) > 0)
odd = N / 2 + 1;
even = N / 2;
// Case 1: 2 odds and 1 even
count = (((odd * (odd - 1)) / 2)
* even) % MOD;
// Case 2: 3 evens
count = (count + ((even * (even
- 1) * (even - 2)) / 6))
% MOD;
return (long)count;
}
// Driver code
static public void main (String[] args)
{
int n = 10;
System.out.println(countWays(n));
}
}
// This code is contributed by vt_m.
Python 3
# Python3 code for above implementation
MOD = 1000000007
# Function to count number of ways
def countWays( N ):
odd = N / 2
if N & 1:
odd = N / 2 + 1
even = N / 2
# Case 1: 2 odds and 1 even
count = (((odd * (odd - 1)) / 2) * even) % MOD
# Case 2: 3 evens
count = (count + ((even * (even - 1) *
(even - 2)) / 6)) % MOD
return count
# Driver code
n = 10
print(int(countWays(n)))
# This code is contributed by "Sharad_Bhardwaj"
C#
// C# program for above implementation
using System;
public class GFG {
static long MOD = 1000000007;
// Function to count number of ways
static long countWays(int N)
{
long count, odd = N / 2, even;
if ((N & 1) > 0)
odd = N / 2 + 1;
even = N / 2;
// Case 1: 2 odds and 1 even
count = (((odd * (odd - 1)) / 2)
* even) % MOD;
// Case 2: 3 evens
count = (count + ((even * (even
- 1) * (even - 2)) / 6))
% MOD;
return (long)count;
}
// Driver code
static public void Main ()
{
int n = 10;
Console.WriteLine(countWays(n));
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program for
// above implementation
$MOD = 1000000007;
// Function to count
// number of ways
function countWays($N)
{
global $MOD;
$count;
$odd =$N / 2;
$even;
if ($N & 1)
$odd = $N / 2 + 1;
$even = $N / 2;
// Case 1: 2 odds
// and 1 even
$count = ((($odd * ($odd - 1)) / 2) *
$even) % $MOD;
// Case 2: 3 evens
$count = ($count + (($even * ($even - 1) *
($even - 2)) / 6)) % $MOD;
return $count;
}
// Driver Code
$n = 10;
echo countWays($n);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// Javascript program for above implementation
let MOD = 1000000007;
// Function to count number of ways
function countWays(N)
{
let count, odd = N / 2, even;
if ((N & 1) > 0)
odd = N / 2 + 1;
even = N / 2;
// Case 1: 2 odds and 1 even
count = (((odd * (odd - 1)) / 2)
* even) % MOD;
// Case 2: 3 evens
count = (count + ((even * (even
- 1) * (even - 2)) / 6))
% MOD;
return count;
}
// Driver code
let n = 10;
document.write(countWays(n));
// This code is contributed by code_hunt.
</script>
输出:
60
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