购买最少的物品而无需找零并提供硬币

您有无限数量的 10 卢比硬币和一枚 r 卢比硬币，您需要购买每件价格为 k 的最低物品，这样您就不会要求找零。
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.

很明显，我们可以支付 10 件物品而无需找零（通过支付所需数量的 10 卢比硬币而不使用 r 卢比硬币）。但也许你可以少买几把锤子，不用找零就付钱。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

Javascript

 
// Javascript program of the above approach
 
function minItems(k, r)
{
     
// See if we can buy less than 10 items
// Using 10 Rs coins and one r Rs coin
for (let 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
 
    let k = 15;
    let r = 2;
    document.write(minItems(k, r));
 

输出：

2