不找零购买最少物品并赠送硬币
你有无限数量的 10 卢比硬币和正好一枚 r 卢比硬币，你需要购买最低成本为 k 的物品，这样你就不会要求零钱。 例:
输入:k = 15，r = 2 输出:2 你应该买两根电缆，付 2*15=30 卢比。很明显，你可以不用找零就付这笔钱。 输入:k = 237，r = 7 输出:1 买一根电缆就够了。
很明显，我们可以支付 10 件物品而不需要任何零钱(通过支付所需数量的 10 卢比硬币，并且不使用 r 卢比硬币)。但也许你可以少买几把锤子，不用找零就能付款。请注意，您应该至少购买一件物品。
C++
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// See if we can buy less than 10 items
// Using 10 Rs coins and one r Rs coin
function minItems($k, $r)
{
for ($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
$k = 15;
$r = 2;
echo minItems($k, $r);
// This code is contributed by Rajput-Ji
?>
java 描述语言
<script>
// 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));
</script>
Output:
2
版权属于：月萌API www.moonapi.com，转载请注明出处
