给定集合中给定函数的最小可能值
给定一个表示有限集合{1，2，3，…的集合 Sn ..，n}。锌表示锡的所有子集的集合，正好包含 2 个元素。任务是找到 F(n)的值。
例:
Input: N = 3
Output: 4
For n=3 we get value 1, 2 times and 2, 1 times
thus the answer would be 1 * 2 + 2 * 1 = 4.
Input: N = 10
Output: 165
对于集合中从左到右的每个值，我们得到每个值， n 值乘以该值的次数。 即对于每个 I，要增加的值为(I+1)*(n–I–1)
以下是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long
// Function to find the value of F(n)
ll findF_N(ll n)
{
ll ans = 0;
for (ll i = 0; i < n; ++i)
ans += (i + 1) * (n - i - 1);
return ans;
}
// Driver code
int main()
{
ll n = 3;
cout << findF_N(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of above approach
import java.io.*;
class GFG {
// Function to find the value of F(n)
static long findF_N(long n)
{
long ans = 0;
for (long i = 0; i < n; ++i)
ans += (i + 1) * (n - i - 1);
return ans;
}
// Driver code
public static void main (String[] args) {
long n = 3;
System.out.println( findF_N(n));
}
}
// This code is contributed by anuj_67..
Python 3
# Python3 implementation of
# above approach
# Function to find the value of F(n)
def findF_N(n):
ans = 0
for i in range(n):
ans = ans + (i + 1) * (n - i - 1)
return ans
# Driver code
n = 3
print(findF_N(n))
# This code is contributed by
# Sanjit_Prasad
C
// C# implementation of above approach
using System;
class GFG
{
// Function to find the
// value of F(n)
static long findF_N(long n)
{
long ans = 0;
for (long i = 0; i < n; ++i)
ans += (i + 1) * (n - i - 1);
return ans;
}
// Driver code
public static void Main ()
{
long n = 3;
Console.WriteLine(findF_N(n));
}
}
// This code is contributed by anuj_67
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// Function to find the value of F(n)
function findF_N($n)
{
$ans = 0;
for ($i = 0; $i < $n; ++$i)
$ans += ($i + 1) * ($n - $i - 1);
return $ans;
}
// Driver code
$n = 3;
echo findF_N($n);
// This code is contributed
// by Akanksha Rai(Abby_akku)
java 描述语言
<script>
// JavaScript implementation of above approach
// Function to find the value of F(n)
function findF_N(n)
{
var ans = 0;
for (var i = 0; i < n; ++i)
ans += (i + 1) * (n - i - 1);
return ans;
}
// Driver code
var n = 3;
document.write( findF_N(n));
</script>
Output:
4
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