购买最低数量的零钱，并给予硬币

您拥有无限数量的10卢比硬币和1卢比硬币，并且您需要购买最低成本为k的物品，以便您不要求找零。

例子：

Input: k = 15, r = 2
Output: 2
You should buy two cables and pay 2*15=30 rupees. It is obvious that you can pay this sum without any change.

Input: k = 237, r = 7
Output:1
It is enough for you to buy one cable.

显然，我们可以支付10件物品而没有任何更改(通过支付所需的10卢比硬币金额，而不使用r卢比硬币)。但是也许您可以购买更少的锤子，而无需支付任何费用。请注意，您应该至少购买一件商品。

C++

#include 
using namespace std;
  
int minItems(int k, int r)
{
   // See if we can buy less than 10 items
   // Using 10 Rs coins and one r Rs coin
   for (int i = 1; i < 10; i++) 
        if ((i * k - r) % 10 == 0 ||
            (i * k) % 10 == 0) 
            return i;
  
    // We can always buy 10 items
    return 10;
}
  
int main()
{
    int k = 15;
    int r = 2;
    cout << minItems(k, r);
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
  
class GFG
{
static int minItems(int k, int r)
{
      
// See if we can buy less than 10 items
// Using 10 Rs coins and one r Rs coin
for (int i = 1; i < 10; i++) 
        if ((i * k - r) % 10 == 0 ||
            (i * k) % 10 == 0) 
            return i;
  
    // We can always buy 10 items
    return 10;
}
  
// Driver Code
public static void main(String args[])
{
    int k = 15;
    int r = 2;
    System.out.println(minItems(k, r));
}
}
  
// This code is contributed 
// by SURENDRA_GANGWAR

Python3

# Python3 implementation of above approach
  
def minItems(k, r) :
  
    # See if we can buy less than 10 items 
    # Using 10 Rs coins and one r Rs coin 
    for i in range(1, 10) : 
            if ((i * k - r) % 10 == 0 or 
                (i * k) % 10 == 0) :
                return i 
      
    # We can always buy 10 items 
    return 10; 
      
# Driver Code
if __name__ == "__main__" : 
  
    k, r = 15 , 2; 
    print(minItems(k, r))
  
# This code is contributed by Ryuga

C#

// C# implementation of above approach
using System;
  
class GFG
{
static int minItems(int k, int r)
{
      
// See if we can buy less than 10 items
// Using 10 Rs coins and one r Rs coin
for (int i = 1; i < 10; i++) 
        if ((i * k - r) % 10 == 0 ||
            (i * k) % 10 == 0) 
            return i;
  
    // We can always buy 10 items
    return 10;
}
  
// Driver Code
public static void Main()
{
    int k = 15;
    int r = 2;
    Console.WriteLine(minItems(k, r));
}
}
  
// This code is contributed 
// by inder_verma

输出：

2