前N个数之和为10的倍数

给定一个整数N ，任务是打印数字总和为10的倍数的前N个项。该系列的前几个词是19、28、37、46、55，…

例子：

Input: N = 5
Output: 19 28 37 46 55

Input: N = 10
Output: 19 28 37 46 55 64 73 82 91 109

方法：可以看出，要获得所需序列的^第N^个项，请找到N的位数之和。如果总和已经是10的倍数，则在N的末尾附加数字0 ，否则在末尾附加最小的可能数字，以使新的数字总和是10的倍数。

For example, to get the 19^th term, since the sum of digits is already a multiple of 10 then append 0 and 190 is the 19^th term of the series.
For N = 5, the minimum digit that can be appended to make the sum of digits as a multiple of 10 is 5 and 55 is the 5^th term of the series.

下面是上述方法的实现：

C++

#include 
using namespace std;
  
const int TEN = 10;
  
// Function to return the
// sum of digits of n
int digitSum(int n)
{
    int sum = 0;
    while (n > 0) {
  
        // Add last digit to the sum
        sum += n % TEN;
  
        // Remove last digit
        n /= TEN;
    }
  
    return sum;
}
  
// Function to return the nth term
// of the required series
int getNthTerm(int n)
{
    int sum = digitSum(n);
  
    // If sum of digit is already
    // a multiple of 10 then append 0
    if (sum % TEN == 0)
        return (n * TEN);
  
    // To store the minimum digit
    // that must be appended
    int extra = TEN - (sum % TEN);
  
    // Return n after appending
    // the required digit
    return ((n * TEN) + extra);
}
  
// Function to print the first n terms
// of the requried series
void firstNTerms(int n)
{
    for (int i = 1; i <= n; i++)
        cout << getNthTerm(i) << " ";
}
  
// Driver code
int main()
{
    int n = 10;
  
    firstNTerms(n);
  
    return 0;
}

Java

// Java implementation of the above approach
class GFG 
{
    final static int TEN = 10; 
      
    // Function to return the 
    // sum of digits of n 
    static int digitSum(int n) 
    { 
        int sum = 0; 
        while (n > 0) 
        { 
      
            // Add last digit to the sum 
            sum += n % TEN; 
      
            // Remove last digit 
            n /= TEN; 
        } 
        return sum; 
    } 
      
    // Function to return the nth term 
    // of the required series 
    static int getNthTerm(int n) 
    { 
        int sum = digitSum(n); 
      
        // If sum of digit is already 
        // a multiple of 10 then append 0 
        if (sum % TEN == 0) 
            return (n * TEN); 
      
        // To store the minimum digit 
        // that must be appended 
        int extra = TEN - (sum % TEN); 
      
        // Return n after appending 
        // the required digit 
        return ((n * TEN) + extra); 
    } 
      
    // Function to print the first n terms 
    // of the requried series 
    static void firstNTerms(int n) 
    { 
        for (int i = 1; i <= n; i++) 
            System.out.print(getNthTerm(i) + " "); 
    } 
      
    // Driver code 
    public static void main (String[] args) 
    { 
        int n = 10; 
      
        firstNTerms(n); 
    } 
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 code for above implementation
TEN = 10
  
# Function to return the
# sum of digits of n
def digitSum(n):
    sum = 0
    while (n > 0):
  
        # Add last digit to the sum
        sum += n % TEN
  
        # Remove last digit
        n //= TEN
  
    return sum
  
# Function to return the nth term
# of the required series
def getNthTerm(n):
    sum = digitSum(n)
  
    # If sum of digit is already
    # a multiple of 10 then append 0
    if (sum % TEN == 0):
        return (n * TEN)
  
    # To store the minimum digit
    # that must be appended
    extra = TEN - (sum % TEN)
  
    # Return n after appending
    # the required digit
    return ((n * TEN) + extra)
  
# Function to print the first n terms
# of the requried series
def firstNTerms(n):
    for i in range(1, n + 1):
        print(getNthTerm(i), end = " ")
  
# Driver code
n = 10
  
firstNTerms(n)
  
# This code is contributed by Mohit Kumar

C#

// C# Program to Find the Unique elements 
// in linked lists
using System;
      
class GFG 
{
    readonly static int TEN = 10; 
      
    // Function to return the 
    // sum of digits of n 
    static int digitSum(int n) 
    { 
        int sum = 0; 
        while (n > 0) 
        { 
      
            // Add last digit to the sum 
            sum += n % TEN; 
      
            // Remove last digit 
            n /= TEN; 
        } 
        return sum; 
    } 
      
    // Function to return the nth term 
    // of the required series 
    static int getNthTerm(int n) 
    { 
        int sum = digitSum(n); 
      
        // If sum of digit is already 
        // a multiple of 10 then append 0 
        if (sum % TEN == 0) 
            return (n * TEN); 
      
        // To store the minimum digit 
        // that must be appended 
        int extra = TEN - (sum % TEN); 
      
        // Return n after appending 
        // the required digit 
        return ((n * TEN) + extra); 
    } 
      
    // Function to print the first n terms 
    // of the requried series 
    static void firstNTerms(int n) 
    { 
        for (int i = 1; i <= n; i++) 
            Console.Write(getNthTerm(i) + " "); 
    } 
      
    // Driver code 
    public static void Main (String[] args) 
    { 
        int n = 10; 
      
        firstNTerms(n); 
    } 
}
  
// This code is contributed by 29AjayKumar

输出：

19 28 37 46 55 64 73 82 91 109