给定一个以十进制为基数的数 N，求任意基数 B 的位数之和
原文:https://www . geesforgeks . org/给定一个十进制数 n-base-find-任何基数中的数字总和-b/
给定一个以十进制为基数的数 N ，任务是求任意基数的数的位数之和 B 。 例:
输入: N = 100，B = 8 输出: 9 解释: (100) 8 = 144 和(144) = 1 + 4 + 4 = 9 输入: N = 50，B = 2 输出: 3 解释:T21
方法:用基数 B 对数字 N 进行模运算，再用 N = N / B 更新数字，找到单位数字，每一步加单位数字更新和。 以下是上述办法的实施情况
C++
// C++ Implementation to Compute Sum of
// Digits of Number N in Base B
#include <iostream>
using namespace std;
// Function to compute sum of
// Digits of Number N in base B
int sumOfDigit(int n, int b)
{
// Initialize sum with 0
int unitDigit, sum = 0;
while (n > 0) {
// Compute unit digit of the number
unitDigit = n % b;
// Add unit digit in sum
sum += unitDigit;
// Update value of Number
n = n / b;
}
return sum;
}
// Driver function
int main()
{
int n = 50;
int b = 2;
cout << sumOfDigit(n, b);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Implementation to Compute Sum of
// Digits of Number N in Base B
class GFG
{
// Function to compute sum of
// Digits of Number N in base B
static int sumOfDigit(int n, int b)
{
// Initialize sum with 0
int unitDigit, sum = 0;
while (n > 0)
{
// Compute unit digit of the number
unitDigit = n % b;
// Add unit digit in sum
sum += unitDigit;
// Update value of Number
n = n / b;
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int n = 50;
int b = 2;
System.out.print(sumOfDigit(n, b));
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 Implementation to Compute Sum of
# Digits of Number N in Base B
# Function to compute sum of
# Digits of Number N in base B
def sumOfDigit(n, b):
# Initialize sum with 0
unitDigit = 0
sum = 0
while (n > 0):
# Compute unit digit of the number
unitDigit = n % b
# Add unit digit in sum
sum += unitDigit
# Update value of Number
n = n // b
return sum
# Driver code
n = 50
b = 2
print(sumOfDigit(n, b))
# This code is contributed by ApurvaRaj
C
// C# Implementation to Compute Sum of
// Digits of Number N in Base B
using System;
class GFG
{
// Function to compute sum of
// Digits of Number N in base B
static int sumOfDigit(int n, int b)
{
// Initialize sum with 0
int unitDigit, sum = 0;
while (n > 0)
{
// Compute unit digit of the number
unitDigit = n % b;
// Add unit digit in sum
sum += unitDigit;
// Update value of Number
n = n / b;
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int n = 50;
int b = 2;
Console.Write(sumOfDigit(n, b));
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// Javascript Implementation to Compute Sum of
// Digits of Number N in Base B
// Function to compute sum of
// Digits of Number N in base B
function sumOfDigit(n, b)
{
// Initialize sum with 0
var unitDigit, sum = 0;
while (n > 0)
{
// Compute unit digit of the number
unitDigit = n % b;
// Add unit digit in sum
sum += unitDigit;
// Update value of Number
n = parseInt(n / b);
}
return sum;
}
// Driver function
var n = 50;
var b = 2;
document.write(sumOfDigit(n, b));
// This code is contributed by rutvik_56.
</script>
Output:
3
时间复杂度: O(N)，N =位数
辅助空间: O(1)
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