一个数字的所有可能的二进制表示形式对的十进制等效总和

给定数字N。任务是查找由给定数字的二进制表示形式形成的所有对的十进制等效值之和。

例子：

Input: N = 4 
Output: 4 
Binary equivalent of 4 is 100. 
All possible pairs are 10, 10, 00 and their decimal equivalent are 2, 2, 0 respectively. 
So, 2 + 2+ 0 = 4

Input: N = 11 
Output: 13 
All possible pairs are: 10, 11, 11, 01, 01, 11 
Sum = 2 + 3 + 3 + 1 + 1 + 3 = 13

方法：

1. 找出N的等价二进制并将其存储在向量中。
2. 运行两个循环，以考虑由向量中存储的二进制等价位构成的每一对。
3. 找到所有对的十进制等效项并将其相加。
4. 返回总和。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to find the sum
int sumOfPairs(int n)
{
 
    // Store the Binary equivalent of decimal
    // number in reverse order
    vector v;
    int sum = 0;
 
    // Calculate binary equivalent of decimal number
    while (n > 0) {
        v.push_back(n % 2);
        n = n / 2;
    }
 
    // for correct binary representation
    reverse(v.begin(), v.end());
 
    // Consider every pair
    for (int i = 0; i < v.size() - 1; i++) {
        for (int j = i + 1; j < v.size(); j++)
 
        {
            // handles all combinations of 01
            if (v[i] == 0 && v[j] == 1)
                sum += 1;
 
            // handles all combinations of 11
            if (v[i] == 1 && v[j] == 1)
                sum += 3;
 
            // handles all combinations of 10
            if (v[i] == 1 && v[j] == 0)
                sum += 2;
        }
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int N = 5;
 
    cout << sumOfPairs(N);
 
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
 
class GFG
{
public static int sumOfPairs(int n)
{
    // Store the Binary equivalent 
    // of decimal number in reverse order
    ArrayList v = new ArrayList();
    int sum = 0;
     
    // Calculate binary equivalent
    // of decimal number
    while (n > 0)
    {
        v.add(n % 2);
        n = n / 2;
}
 
Collections.reverse(v);
     
for (int i = 0; i < v.size() - 1; i++)
{
    for (int j = i + 1; j < v.size(); j++)
 
    {
        // handles all combinations of 01
        if (v.get(i) == 0 && v.get(j) == 1)
            sum += 1;
 
        // handles all combinations of 11
        if (v.get(i) == 1 && v.get(j) == 1)
            sum += 3;
 
        // handles all combinations of 10
        if (v.get(i) == 1 && v.get(j) == 0)
            sum += 2;
    }
}
 
return sum;
}
 
// Driver Code
public static void main (String[] args)
{
    int N = 5;
 
    System.out.print(sumOfPairs(N));
}
}
 
// This code is contributed by Kirti_Mangal

Python3

# Python3 program to find the sum
 
# Function to find the sum
def sumofPairs(n) :
 
    # Store the Binary equivalent of decimal
    # number in reverse order
    v = []
    sum = 0
 
    # Calculate binary equivalent of decimal number
    while n > 0 :
        v.append(n % 2)
        n = n // 2
 
    # for correct binary representation
    v.reverse()
 
    # Consider every pair
    for i in range(len(v) - 1) :
 
        for j in range(i + 1, len(v)) :
 
            # handles all combinations of 01
            if v[i] == 0 and v[j] == 1 :
                sum += 1
 
            #  handles all combinations of 11
            if v[i] == 1 and v[j] == 1 :
                sum += 3
 
            # handles all combinations of 10
            if v[i] == 1 and v[j] == 0 :
                sum += 2
 
    return sum
 
# Driver Code
if __name__ == "__main__" :
     
    N = 5
 
    # function calling
    print(sumofPairs(N))
 
# This code is contributed by ANKITRAI1

C#

// C# implementation of above approach
using System;
using System.Collections.Generic;
 
class GFG
{
    public static int sumOfPairs(int n)
    {
        // Store the Binary equivalent
        // of decimal number in reverse order
        List v = new List();
        int sum = 0;
 
        // Calculate binary equivalent
        // of decimal number
        while (n > 0)
        {
            v.Add(n % 2);
            n = n / 2;
        }
 
    v.Reverse();
 
    for (int i = 0; i < v.Count - 1; i++)
    {
        for (int j = i + 1; j < v.Count; j++)
 
        {
            // handles all combinations of 01
            if (v[i] == 0 && v[j] == 1)
                sum += 1;
 
            // handles all combinations of 11
            if (v[i] == 1 && v[j] == 1)
                sum += 3;
 
            // handles all combinations of 10
            if (v[i] == 1 && v[j] == 0)
                sum += 2;
        }
    }
 
    return sum;
    }
 
    // Driver Code
    public static void Main (String[] args)
    {
        int N = 5;
 
        Console.WriteLine(sumOfPairs(N));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// Javascript implementation of above approach
 
// Function to find the sum
function sumOfPairs(n)
{
     
    // Store the Binary equivalent of
    // decimal number in reverse order
    var v = [];
    var sum = 0;
 
    // Calculate binary equivalent
    // of decimal number
    while (n > 0)
    {
        v.push(n % 2);
        n = parseInt(n / 2);
    }
 
    // For correct binary representation
    v.reverse();
 
    // Consider every pair
    for(var i = 0; i < v.length - 1; i++)
    {
        for(var j = i + 1; j < v.length; j++)
        {
             
            // Handles all combinations of 01
            if (v[i] == 0 && v[j] == 1)
                sum += 1;
 
            // Handles all combinations of 11
            if (v[i] == 1 && v[j] == 1)
                sum += 3;
 
            // Handles all combinations of 10
            if (v[i] == 1 && v[j] == 0)
                sum += 2;
        }
    }
    return sum;
}
 
// Driver code
var N = 5;
document.write(sumOfPairs(N));
 
// This code is contributed by rrrtnx
 

输出：

6

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。