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📜  以2为基数到N / 2的所有底数的N个数字的总和

📅  最后修改于: 2021-06-26 08:47:49             🧑  作者: Mango

给定一个整数N ,任务是查找从2N / 2的所有基数中写入的N的数字的总和。
例子:

方法:

  • 对于从2(n / 2)的每个底数,使用以下公式计算特定底数中n的数字:
    • 通过将n除以底数来计算余数,余数是该底数中n的数字之一。
    • 将数字加到总和并更新n为(n = n / base)
    • n> 0时重复上述步骤
  • 打印在先前步骤中计算出的总和

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// Function to calculate the sum of the digits of
// n in the given base
int solve(int n, int base)
{
    // Sum of digits
    int sum = 0;
 
    while (n > 0) {
 
        // Digit of n in the given base
        int remainder = n % base;
 
        // Add the digit
        sum += remainder;
        n = n / base;
    }
 
    return sum;
}
 
// Function to calculate the sum of
// digits of n in bases from 2 to n/2
void SumsOfDigits(int n)
{
    // to store digit sum in all bases
    int sum = 0;
 
    // function call for multiple bases
    for (int base = 2; base <= n / 2; ++base)
        sum += solve(n, base);
 
    cout << sum;
}
 
// Driver program
int main()
{
    int n = 8;
    SumsOfDigits(n);
    return 0;
}


Java
// Java implementation of the approach
 
import java.io.*;
 
class GFG {
     
 
// Function to calculate the sum of the digits of
// n in the given base
static int solve(int n, int base)
{
    // Sum of digits
    int sum = 0;
 
    while (n > 0) {
 
        // Digit of n in the given base
        int remainder = n % base;
 
        // Add the digit
        sum += remainder;
        n = n / base;
    }
 
    return sum;
}
 
// Function to calculate the sum of
// digits of n in bases from 2 to n/2
static void SumsOfDigits(int n)
{
    // to store digit sum in all bases
    int sum = 0;
 
    // function call for multiple bases
    for (int base = 2; base <= n / 2; ++base)
        sum += solve(n, base);
 
    System.out.println(sum);
}
 
// Driver program
 
 
    public static void main (String[] args) {
        int n = 8;
    SumsOfDigits(n);
    }
     
}
// This code is contributed by anuj_67..


Python3
# Python 3 implementation of the approach
from math import floor
# Function to calculate the sum of the digits of
# n in the given base
def solve(n, base):
    # Sum of digits
    sum = 0
 
    while (n > 0):
        # Digit of n in the given base
        remainder = n % base
 
        # Add the digit
        sum = sum + remainder
        n = int(n / base)
     
    return sum
 
# Function to calculate the sum of
# digits of n in bases from 2 to n/2
def SumsOfDigits(n):
     
    # to store digit sum in all base
    sum = 0
    N = floor(n/2)
    # function call for multiple bases
    for base in range(2,N+1,1):
        sum = sum + solve(n, base)
 
    print(sum)
 
# Driver program
if __name__ == '__main__':
    n = 8
    SumsOfDigits(n)
     
# This code is contributed by
# Surendra_Gangwar


C#
// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to calculate the sum of
// the digits of n in the given base
static int solve(int n, int base1)
{
    // Sum of digits
    int sum = 0;
 
    while (n > 0)
    {
 
        // Digit of n in the given base
        int remainder1 = n % base1;
 
        // Add the digit
        sum += remainder1;
        n = n / base1;
    }
 
    return sum;
}
 
// Function to calculate the sum of
// digits of n in base1s from 2 to n/2
static void SumsOfDigits(int n)
{
    // to store digit sum in all bases
    int sum = 0;
 
    // function call for multiple bases
    for (int base1 = 2;
             base1 <= n / 2; ++base1)
        sum += solve(n, base1);
 
    Console.WriteLine(sum);
}
 
// Driver Code
public static void Main (String []args)
{
    int n = 8;
    SumsOfDigits(n);
}
}
 
// This code is contributed by Arnab Kundu


PHP
 0)
    {
 
        // Digit of n in the given base
        $remainder = $n % $base;
 
        // Add the digit
        $sum += $remainder;
        $n = $n / $base;
    }
 
    return $sum;
}
 
// Function to calculate the sum of
// digits of n in bases from 2 to n/2
function SumsOfDigits($n)
{
    // to store digit sum in all bases
    $sum = 0;
 
    // function call for multiple bases
    for ($base = 2;
         $base <= $n / 2; ++$base)
        $sum += solve($n, $base);
 
    echo $sum;
}
 
// Driver Code
$n = 8;
SumsOfDigits($n);
 
// This code is contributed
// by Akanksha Rai
?>


Javascript


输出:
7

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