使用两个方程式找到重复和缺失的数字

给定大小为N的数组arr [] ，范围[1，N]中的每个整数仅出现一次，除了A出现两次而B丢失了。任务是找到数字A和B。
例子：

Input: arr[] = {1, 2, 2, 3, 4} 
Output: 
A = 2 
B = 5
Input: arr[] = {5, 3, 4, 1, 1} 
Output: 
A = 1 
B = 2 
 

方法：根据前N个自然数的总和，

SumN = 1 + 2 + 3 + … + N = (N * (N + 1)) / 2 
And, let the sum of all the array elements be Sum. Now, 
SumN = Sum – A + B 
A – B = Sum – SumN …(equation 1) 
 

根据前N个自然数的平方和

SumSqN = 1² + 2² + 3² + … + N² = (N * (N + 1) * (2 * n + 1)) / 6 
And, let the sum of the squares of all the array elements be SumSq. Now, 
SumSq = SumSqN + A² – B² 
SumSq – SumSqN = (A + B) * (A – B) …(equation 2) 
 

将等式2中等式1的(A – B)值
SumSq – SumSqN =(A + B)*(Sum – SumN)
A + B =(SumSq – SumSqN)/(Sum – SumN)…(等式3)
求解方程式1和方程式3将得出，

B = (((SumSq – SumSqN) / (Sum – SumN)) + SumN – Sum) / 2 
And, A = Sum – SumN + B 
 

下面是上述方法的实现：

C++

//C++ implementation of the approach
 
#include 
#include
#include 
 
using namespace std;
 
    // Function to print the required numbers
 void findNumbers(int arr[], int n)
    {
 
        // Sum of first n natural numbers
        int sumN = (n * (n + 1)) / 2;
 
        // Sum of squares of first n natural numbers
        int sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;
 
        // To store the sum and sum of squares
        // of the array elements
        int sum = 0, sumSq = 0, i;
 
        for (i = 0; i < n; i++) {
            sum += arr[i];
            sumSq = sumSq + (pow(arr[i], 2));
        }
 
        int B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
        int A = sum - sumN + B;
         cout << "A = " ;
         cout << A << endl;
         cout << "B = " ;
         cout << B << endl;
    }
 
    // Driver code
int main() {
        int arr[] = { 1, 2, 2, 3, 4 };
        int n = sizeof(arr)/sizeof(arr[0]);
        findNumbers(arr, n);
    return 0;
}

Java

// Java implementation of the approach
public class GFG {
 
    // Function to print the required numbers
    static void findNumbers(int arr[], int n)
    {
 
        // Sum of first n natural numbers
        int sumN = (n * (n + 1)) / 2;
 
        // Sum of squares of first n natural numbers
        int sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;
 
        // To store the sum and sum of squares
        // of the array elements
        int sum = 0, sumSq = 0, i;
 
        for (i = 0; i < n; i++) {
            sum += arr[i];
            sumSq += Math.pow(arr[i], 2);
        }
 
        int B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
        int A = sum - sumN + B;
        System.out.println("A = " + A + "\nB = " + B);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 2, 3, 4 };
        int n = arr.length;
 
        findNumbers(arr, n);
    }
}

Python3

# Python3 implementation of the approach
 
import math
# Function to print the required numbers
def findNumbers(arr, n):
     
 
        # Sum of first n natural numbers
        sumN = (n * (n + 1)) / 2;
 
        # Sum of squares of first n natural numbers
        sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;
 
        # To store the sum and sum of squares
        # of the array elements
        sum = 0;
        sumSq = 0;
 
        for i in range(0,n):
            sum = sum + arr[i];
            sumSq = sumSq + (math.pow(arr[i], 2));
         
 
        B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
        A = sum - sumN + B;
        print("A = ",int(A)) ;
        print("B = ",int(B));
     
 
# Driver code
 
arr = [ 1, 2, 2, 3, 4 ];
n = len(arr);
findNumbers(arr, n);
 
#This code is contributed by Shivi_Aggarwal

C#

// C# implementation of the approach
using System;
public class GFG {
 
    // Function to print the required numbers
    static void findNumbers(int []arr, int n)
    {
 
        // Sum of first n natural numbers
        int sumN = (n * (n + 1)) / 2;
 
        // Sum of squares of first n natural numbers
        int sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;
 
        // To store the sum and sum of squares
        // of the array elements
        int sum = 0, sumSq = 0, i;
 
        for (i = 0; i < n; i++) {
            sum += arr[i];
            sumSq += (int)Math.Pow(arr[i], 2);
        }
 
        int B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
        int A = sum - sumN + B;
        Console.WriteLine("A = " + A + "\nB = " + B);
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = { 1, 2, 2, 3, 4 };
        int n = arr.Length;
 
        findNumbers(arr, n);
    }
}
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript implementation of the approach
 
// Function to print the required numbers
function findNumbers(arr, n)
{
 
    // Sum of first n natural numbers
    sumN = (n * (n + 1)) / 2;
 
    // Sum of squares of first n
    // natural numbers
    sumSqN = (n * (n + 1) *
                (2 * n + 1)) / 6;
 
    // To store the sum and sum of
    // squares of the array elements
    let sum = 0 ;
    let sumSq = 0 ;
 
    for (let i = 0;i < n; i++)
    {
        sum += arr[i];
        sumSq += Math.pow(arr[i], 2);
    }
 
    B = (((sumSq - sumSqN) / (sum - sumN)) +
                                sumN - sum) / 2;
    A = sum - sumN + B;
    document.write( "A = "+ A, "<br>B = ", B);
}
 
// Driver code
let arr = [ 1, 2, 2, 3, 4 ];
n = arr.length ;
 
findNumbers(arr, n);
 
// This code is contributed
// by bobby
 

输出：

A = 2
B = 5

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