最小整数，其数字和为M且为N的倍数

给定两个正整数N和M，任务是找到可被N整除且数字总和为M的最小正整数。如果int范围内不存在这样的整数，则打印-1 。

例子：

Input: N = 13, M = 32
Output: 8879
8879 is divisible by 13 and its 
Sum of digits of 8879 is 8+8+7+9 = 32 
i.e. equals to M

Input: N = 8, M = 32;
Output: 8888

方法：从N开始，遍历n的所有倍数，然后检查其数字总和是否等于m。此问题可以通过log _n (INT_MAX)* log ₁₀ (INT_MAX)时间解决。通过以至少m / 9位数的数字开始迭代，可以提高此方法的效率。

下面是该方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Function to return digit sum
int digitSum(int n)
{
    int ans = 0;
    while (n) {
        ans += n % 10;
        n /= 10;
    }
  
    return ans;
}
  
// Function to find out the smallest integer
int findInt(int n, int m)
{
    int minDigit = floor(m / 9);
  
    // Start of the iterator (Smallest multiple of n)
    int start = pow(10, minDigit) - 
                (int)pow(10, minDigit) % n;
  
    while (start < INT_MAX) {
        if (digitSum(start) == m)
            return start;
        else
            start += n;
    }
    return -1;
}
  
// Driver code
int main()
{
    int n = 13, m = 32;
    cout << findInt(n, m);
    return 0;
}

Java

// Java implementation of the above approach 
  
class GFG
{
  
    // Function to return digit sum 
    static int digitSum(int n) 
    { 
        int ans = 0; 
        while (n != 0)
        { 
            ans += n % 10; 
            n /= 10; 
        } 
      
        return ans; 
    } 
      
    // Function to find out the 
    // smallest integer 
    static int findInt(int n, int m) 
    { 
        int minDigit = (int)Math.floor((double)(m / 9)); 
      
        // Start of the iterator (Smallest multiple of n) 
        int start = (int)Math.pow(10, minDigit) - 
                    (int)Math.pow(10, minDigit) % n; 
      
        while (start < Integer.MAX_VALUE) 
        { 
            if (digitSum(start) == m) 
                return start; 
            else
                start += n; 
        } 
        return -1; 
    } 
  
    // Driver code 
    static public void main(String args[]) 
    { 
        int n = 13, m = 32; 
        System.out.print(findInt(n, m)); 
    } 
}
  
// This code is contributed 
// by Akanksha Rai

Python3

# Python 3 implementation of the 
# above approach
from math import floor, pow
  
import sys
  
# Function to return digit sum
def digitSum(n):
    ans = 0;
    while (n):
        ans += n % 10;
        n = int(n / 10);
  
    return ans
  
# Function to find out the smallest 
# integer
def findInt(n, m):
    minDigit = floor(m / 9)
  
    # Start of the iterator (Smallest 
    # multiple of n)
    start = (int(pow(10, minDigit)) - 
             int(pow(10, minDigit)) % n)
  
    while (start < sys.maxsize):
        if (digitSum(start) == m):
            return start
        else:
            start += n
    return -1
  
# Driver code
if __name__ == '__main__':
    n = 13
    m = 32
    print(findInt(n, m))
  
# This code is cotributed by
# Surendra_Gangwar

C#

// C# implementation of the above approach 
using System;
  
class GFG
{
  
    // Function to return digit sum 
    static int digitSum(int n) 
    { 
        int ans = 0; 
        while (n != 0)
        { 
            ans += n % 10; 
            n /= 10; 
        } 
      
        return ans; 
    } 
      
    // Function to find out the 
    // smallest integer 
    static int findInt(int n, int m) 
    { 
        int minDigit = (int)Math.Floor((double)(m / 9)); 
      
        // Start of the iterator (Smallest multiple of n) 
        int start = (int)Math.Pow(10, minDigit) - 
                    (int)Math.Pow(10, minDigit) % n; 
      
        while (start < int.MaxValue) 
        { 
            if (digitSum(start) == m) 
                return start; 
            else
                start += n; 
        } 
        return -1; 
    } 
  
    // Driver code 
    static public void Main() 
    { 
        int n = 13, m = 32; 
        Console.WriteLine(findInt(n, m)); 
    } 
}
  
// This code is contributed by Ryuga

输出：

8879