检查一个数字是否可以表示为至少一位数字等于 K 的数字之和

给定整数N和K ，任务是检查一个数字是否可以表示为至少有一位数字等于 K 的数字的总和。

例子：

Input: N = 68, K = 7
Output: YES
Explanation: 68 = (27 + 17 + 17 + 7). Each number has atleast one digit equal to 7.

Input: N = 23, K = 3
Output: YES
Explanation: 23 itself contains a digit equal to 3.

方法：给定的问题可以通过使用简单的数学概念来解决。请按照以下步骤解决问题：

• 将变量temp初始化为k ，并取一个计数器说count ，将其分配为 0
• 迭代直到temp和N的最后一位不相等，并且在每次迭代中
  • 将temp的值增加k
  • 保持迭代计数并在计数大于 10 时中断循环
• 检查temp和N的最后一位是否相等，以及temp <= N的值是否相等：
  • 如果满足上述条件，则返回 true
  • 否则返回假
• 此外，如果k * 10 <= N ，则返回 true
• 否则返回假

下面是上述方法的实现：

C++

// C++ implementation for the above approach
 
#include 
using namespace std;
 
// Function to Check if a number can
// be equal to sum of numbers having
// at least one digit equal to k
bool checkEqualtoSum(int N, int k)
{
    // Temporary variable to
    // store k
    int temp = k;
     
    // Variable for count
    int count = 0;
     
    // Iterating till count is less or
    // equal to 10 and N % 10 is not
    // equal to temp % 10
    while(count <= 10 &&
                  N % 10 != temp % 10) {
       
        temp += k;
        count++;
    }
     
    // If N % 10 is equal to temp % 10
    // and temp is less or equal to N,
    // return true
    if(N % 10 == temp % 10 && temp <= N)
        return true;
     
    // If k * 10 <= N, return true
    if(k * 10 <= N)
        return true;
     
    // Else return false
    return false;
}
 
// Driver Code
int main()
{
    int N = 68;
    int K = 7;
 
      // Call the function
    if(checkEqualtoSum(N, K))
          cout << "YES";
    else cout << "NO";
     
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG {
 
    // Function to Check if a number can
    // be equal to sum of numbers having
    // at least one digit equal to k
    static boolean checkEqualtoSum(int N, int k)
    {
        // Temporary variable to
        // store k
        int temp = k;
 
        // Variable for count
        int count = 0;
 
        // Iterating till count is less or
        // equal to 10 and N % 10 is not
        // equal to temp % 10
        while (count <= 10 && N % 10 != temp % 10) {
 
            temp += k;
            count++;
        }
 
        // If N % 10 is equal to temp % 10
        // and temp is less or equal to N,
        // return true
        if (N % 10 == temp % 10 && temp <= N)
            return true;
 
        // If k * 10 <= N, return true
        if (k * 10 <= N)
            return true;
 
        // Else return false
        return false;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given Input
        int N = 68;
        int K = 7;
 
        // Call the function
        if (checkEqualtoSum(N, K))
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}
 
// This code is contributed by dwivediyash

Python3

# python implementation for the above approach
 
# Function to Check if a number can
# be equal to sum of numbers having
# at least one digit equal to k
 
 
def checkEqualtoSum(N, k):
 
    # Temporary variable to
    # store k
    temp = k
 
    # Variable for count
    count = 0
 
    # Iterating till count is less or
    # equal to 10 and N % 10 is not
    # equal to temp % 10
    while(count <= 10 and N % 10 != temp % 10):
 
        temp += k
        count += 1
 
    # If N % 10 is equal to temp % 10
    # and temp is less or equal to N,
    # return true
    if(N % 10 == temp % 10 and temp <= N):
        return True
 
    # If k * 10 <= N, return true
    if(k * 10 <= N):
        return True
 
    # Else return false
    return False
 
 
# Driver Code
if __name__ == "__main__":
 
    N = 68
    K = 7
 
    # Call the function
    if(checkEqualtoSum(N, K)):
        print("YES")
    else:
        print("NO")
 
    # This code is contributed by rakeshsahni

Javascript

    // JavaScript implementation for the above approach
 
    // Function to Check if a number can
    // be equal to sum of numbers having
    // at least one digit equal to k
    const checkEqualtoSum = (N, k) => {
        // Temporary variable to
        // store k
        let temp = k;
 
        // Variable for count
        let count = 0;
 
        // Iterating till count is less or
        // equal to 10 and N % 10 is not
        // equal to temp % 10
        while (count <= 10 &&
            N % 10 != temp % 10) {
 
            temp += k;
            count++;
        }
 
        // If N % 10 is equal to temp % 10
        // and temp is less or equal to N,
        // return true
        if (N % 10 == temp % 10 && temp <= N)
            return true;
 
        // If k * 10 <= N, return true
        if (k * 10 <= N)
            return true;
 
        // Else return false
        return false;
    }
 
    // Driver Code
    let N = 68;
    let K = 7;
 
    // Call the function
    if (checkEqualtoSum(N, K))
        document.write("YES");
    else document.write("NO");
 
    // This code is contributed by rakeshsahni
 

C#

// C# implementation for the above approach
using System;
class gFG
{
   
    // Function to Check if a number can
    // be equal to sum of numbers having
    // at least one digit equal to k
    static bool checkEqualtoSum(int N, int k)
    {
        // Temporary variable to
        // store k
        int temp = k;
 
        // Variable for count
        int count = 0;
 
        // Iterating till count is less or
        // equal to 10 and N % 10 is not
        // equal to temp % 10
        while (count <= 10 && N % 10 != temp % 10) {
 
            temp += k;
            count++;
        }
 
        // If N % 10 is equal to temp % 10
        // and temp is less or equal to N,
        // return true
        if (N % 10 == temp % 10 && temp <= N)
            return true;
 
        // If k * 10 <= N, return true
        if (k * 10 <= N)
            return true;
 
        // Else return false
        return false;
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 68;
        int K = 7;
 
        // Call the function
        if (checkEqualtoSum(N, K))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}
 
// This code is contributed by ukasp.

输出

YES

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