给定集合中给定函数的最小值

给定一个代表有限集{1,2,3，…..，n}的集合Sn 。 Zn表示Sn的所有子集的集合，该子集恰好包含2个元素。任务是找到F(n)的值。

$F(n) = \sum_{X\in Z_n }min(X)$

例子：

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

For each value as we go from left to right in the set we get each value, n-value number of times that value.
i.e. For each i, value to be added is (i + 1) * (n – i – 1)

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
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

// 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..

Python3

# 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

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输出：

4