从n个和方程中找到n个变量，其中一个缺失

您将得到一个由n个值组成的数组a []作为a [1]，a [2]…a [n]，它们是n变量中n方程的一部分，其中等式如下：
等式1 => X2 + X3 +….. + Xn = a1
等式2 => X1 + X3 +….. + Xn = a2
等式=> X1 + X2 +…+ X _i-1 + X _{i + 1} .. + Xn = ai
等式=> X1 + X2 +….. + X _n-1 = an

当您在n个变量中具有n个等式时，找到所有变量(X1，X2…Xn)的值。

例子 ：

Input : a[] = {4, 4, 4, 4, 4}
Output : X1 = 1
         X2 = 1
         X3 = 1
         X4 = 1
         X5 = 1

Input : a[] = {2, 5, 6, 4, 8}
Output : X1 = 4.25
         X2 = 1.25
         X3 = 0.25
         X4 = 2.25
         X5 = -1.75

方法：
令X1 + X2 + X3 + …. Xn = SUM
使用值SUM，我们的方程式为

SUM - X1 = a1 -----(1)
SUM - X2 = a2  -----(2)
SUM - Xi = ai  -----(i)
SUM - Xn = an -------(n)
---------------------------------------------------------------
Now, if we add all these equation we will have an equation as :
n*SUM -(X1+X2+...Xn) = a1 + a2 + ...an
n*SUM - SUM = a1 + a2 + ...an
SUM = (a1+a2+...an)/(n-1)

Calculate SUM from above equation.
putting value of SUM in (i), (ii).... we have
X1 = SUM - a1
X2 = SUM - a2

解决方案 ：

X1 = SUM - a1
X2 = SUM - a2
Xi = SUM - ai
Xn = SUM - an

C++

// CPP program to find n-variables
#include 
using namespace std;
  
// function to print n-variable values
void findVar(int a[], int n)
{    
    // calculate value of array SUM
    float SUM = 0;
    for (int i = 0; i < n; i++)
        SUM += a[i];
  
    // Every variable contributes n-1
    // times to sum. So dividing by 
    // n-1 to get sum of all.
    SUM /= (n - 1);
  
    // print the values of n-variables
    for (int i = 0; i < n; i++)
        cout << "X" << (i + 1) 
             << " = " << SUM - a[i] << endl;
}
  
// driver program
int main()
{
    int a[] = { 2, 5, 6, 4, 8 };
    int n = sizeof(a) / sizeof(a[0]);
    findVar(a, n);
    return 0;
}

Java

// Java program to 
// find n-variables
import java.io.*;
  
class GFG
{
    // function to print
    // n-variable values
    static void findVar(int []a, 
                        int n)
    { 
        // calculate value+
        // of array SUM
        float SUM = 0;
        for (int i = 0; i < n; i++)
            SUM += a[i];
      
        // Every variable contributes 
        // n-1 times to sum. So dividing 
        // by n-1 to get sum of all.
        SUM /= (n - 1);
      
        // print the values
        // of n-variables
        for (int i = 0; i < n; i++)
            System.out.print("X" + (i + 1) + 
                             " = " + (SUM - 
                             a[i]) + "\n");
    }
      
    // Driver Code
    public static void main(String args[])
    {
        int []a = new int[]{2, 5, 6, 4, 8};
        int n = a.length;
        findVar(a, n);
    }
}
  
// This code is contributed by 
// Manish Shaw(manishshaw1)

Python3

# Python3 program to 
# find n-variables
  
# function to print 
# n-variable values
def findVar(a, n):
      
    # calculate value of 
    # array SUM
    SUM = 0;
    for i in range(n):
        SUM += a[i];
  
    # Every variable contributes 
    # n-1 times to sum. So  
    # dividing by n-1 to get sum 
    # of all.
    SUM = SUM / (n - 1);
  
    # print the values 
    # of n-variables
    for i in range(n):
        print("X" , (i + 1), 
              " = ", SUM - a[i]);
  
# Driver Code
a = [2, 5, 6, 4, 8];
n = len(a);
findVar(a, n);
      
# This code is contributed 
# by mits

C#

// C# program to find n-variables
using System;
using System.Linq;
using System.Collections.Generic;
  
class GFG
{
    // function to print
    // n-variable values
    static void findVar(int []a, 
                        int n)
    { 
        // calculate value+
        // of array SUM
        float SUM = 0;
        for (int i = 0; i < n; i++)
            SUM += a[i];
      
        // Every variable contributes 
        // n-1 times to sum. So dividing 
        // by n-1 to get sum of all.
        SUM /= (n - 1);
      
        // print the values
        // of n-variables
        for (int i = 0; i < n; i++)
            Console.Write("X" + (i + 1) + 
                   " = " + (SUM - a[i]) + "\n");
    }
      
    // Driver Code
    static void Main()
    {
        int []a = { 2, 5, 6, 4, 8 };
        int n = a.Length;
        findVar(a, n);
    }
}
  
// This code is contributed by 
// Manish Shaw(manishshaw1)

输出 ：

X1 = 4.25
X2 = 1.25
X3 = 0.25
X4 = 2.25
X5 = -1.75

时间复杂度： O(n)
辅助空间： O(1)