查找一个数字的数字总和，直到总和成为一位数字

本文概述

• C ++
• Java
• python
• C#
• 的PHP
• CPP
• Java
• Python3
• C#
• 的PHP

给定一个数字n，我们需要找出它的数字和，使：

If n < 10    
    digSum(n) = n
Else         
    digSum(n) = Sum(digSum(n))

例子 ：

Input : 1234
Output : 1
Explanation : The sum of 1+2+3+4 = 10, digSum(x) == 10
              Hence ans will be 1+0 = 1

Input : 5674
Output : 4

蛮力方法是对所有数字求和，直到sum < 10。

流程图：

下面是蛮力程序, 用于找到总和。

C ++

// C++ program to find sum of
// digits of a number until
// sum becomes single digit.
#include<bits/stdc++.h>
   
using namespace std;
  
int digSum( int n)
{
     int sum = 0;
     
     // Loop to do sum while
     // sum is not less than
     // or equal to 9
     while (n > 0 || sum > 9)
     {
         if (n == 0)
         {
             n = sum;
             sum = 0;
         }
         sum += n % 10;
         n /= 10;
     }
     return sum;
}
  
// Driver program to test the above function
int main()
{
     int n = 1234;
     cout << digSum(n);
     return 0;
}

Java

// Java program to find sum of
// digits of a number until
// sum becomes single digit.
import java.util.*;
  
public class GfG {
      
     static int digSum( int n)
     {
         int sum = 0 ;
  
         // Loop to do sum while
         // sum is not less than
         // or equal to 9
         while (n > 0 || sum > 9 ) 
         {
             if (n == 0 ) {
                 n = sum;
                 sum = 0 ;
             }
             sum += n % 10 ;
             n /= 10 ;
         }
         return sum;
     }
      
     // Driver code
     public static void main(String argc[])
     {
         int n = 1234 ;
         System.out.println(digSum(n));
     }
}
  
// This code is contributed by Gitanjali.

python

# Python program to find sum of
# digits of a number until
# sum becomes single digit.
import math 
  
# method to find sum of digits 
# of a number until sum becomes 
# single digit
def digSum( n):
     sum = 0
      
     while (n > 0 or sum > 9 ):
      
         if (n = = 0 ):
             n = sum
             sum = 0
          
         sum + = n % 10
         n / = 10
      
     return sum
  
# Driver method
n = 1234
print (digSum(n))
  
# This code is contributed by Gitanjali.

C#

// C# program to find sum of
// digits of a number until
// sum becomes single digit.
using System;
  
class GFG {
      
     static int digSum( int n)
     {
         int sum = 0;
  
         // Loop to do sum while
         // sum is not less than
         // or equal to 9
         while (n > 0 || sum > 9) 
         {
             if (n == 0)
             {
                 n = sum;
                 sum = 0;
             }
             sum += n % 10;
             n /= 10;
         }
         return sum;
     }
      
     // Driver code
     public static void Main()
     {
         int n = 1234;
         Console.Write(digSum(n));
     }
}
  
// This code is contributed by nitin mittal

的PHP

<?php
// PHP program to find sum of
// digits of a number until
// sum becomes single digit.
  
function digSum( $n )
{
     $sum = 0;
      
     // Loop to do sum while
     // sum is not less than
     // or equal to 9
     while ( $n > 0 || $sum > 9)
     {
         if ( $n == 0)
         {
             $n = $sum ;
             $sum = 0;
         }
         $sum += $n % 10;
         $n = (int) $n / 10;
     }
     return $sum ;
}
  
// Driver Code
$n = 1234;
echo digSum( $n );
  
// This code is contributed
// by aj_36
?>

输出：

1

存在一个简单而优雅的O(1)解决方案为此。答案很简单：-

If n == 0
   return 0;

If n % 9 == 0      
    digSum(n) = 9
Else               
    digSum(n) = n % 9

以上逻辑如何运作？

如果数字n可以被9整除, 那么直到位数变成一位数字之前, 其位数的总和始终为9。例如,

设n = 2880

数字总和= 2 + 8 + 8 = 18：18 = 1 + 8 = 9

一个数字可以是9x或9x + k的形式。对于第一种情况, 答案始终为9。对于第二种情况, 答案始终为k。

下面是上述想法的实现：

CPP

#include<bits/stdc++.h> 
using namespace std;
  
int digSum( int n)
{
     if (n == 0) 
        return 0;
     return (n % 9 == 0) ? 9 : (n % 9);
}
  
// Driver program to test the above function
int main()
{
     int n = 9999;
     cout<<digSum(n);
     return 0;
}

Java

import java.io.*;
  
class GFG {
  
     static int digSum( int n)
     {
         if (n == 0 ) 
         return 0 ;
         return (n % 9 == 0 ) ? 9 : (n % 9 );
     }
      
     // Driver program to test the above function
     public static void main (String[] args)
     {
         int n = 9999 ;
         System.out.println(digSum(n));
     }
}
  
// This code is contributed by anuj_67.

Python3

def digSum(n):
  
     if (n = = 0 ):
         return 0
     if (n % 9 = = 0 ):
         return 9 
     else :
         (n % 9 )
  
# Driver program to test the above function
n = 9999
print (digSum(n))
  
# This code is contributed by
# Smitha Dinesh Semwal

C#

using System;
  
class GFG
{
     static int digSum( int n)
     {
         if (n == 0) 
         return 0;
         return (n % 9 == 0) ? 9 : (n % 9);
     }
      
     // Driver Code
     public static void Main ()
     {
         int n = 9999;
         Console.Write(digSum(n));
      
     }
}
  
// This code is contributed by aj_36

的PHP

<?php
  
function digSum( $n )
{
     if ( $n == 0) 
         return 0;
     return ( $n % 9 == 0) ? 9 : ( $n % 9);
}
  
// Driver program to test the above function
$n = 9999;
echo digSum( $n );
  
//This code is contributed by anuj_67.
?>

输出如下：

9

相关文章：

如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。