找出需要支付总金额的最小硬币数
给定总金额 N 和无限数量的价值 1 、 10 和 25 币的硬币。找出准确支付金额所需的最小硬币数量 N 。 举例:
Input : N = 14
Output : 5
You will use *one* coin of value 10 and
*four* coins of value 1.
Input : N = 88
Output : 7
进场:T2【有三种不同情况:
- 如果NT2【10】的值，那么价值为 1 的硬币只能用于支付。
-
当 N > 9 和< 25 时，那么价值为 1 和 10 的硬币将用于支付。在这里，为了尽量减少硬币的使用数量，价值 10 的硬币将是首选。
-
时NT2【24】。那么所有价值 1、10 和 25 的硬币都将用于支付。为了最大限度地减少硬币数量，首要目标是尽可能多地首先使用价值 25 的硬币，然后使用价值 10 的硬币，然后使用价值 1 的硬币。
以下是上述方法的实现:
C++
// C++ program to find the minimum number
// of coins required
#include <iostream>
using namespace std;
// Function to find the minimum number
// of coins required
int countCoins(int n)
{
int c = 0;
if (n < 10) {
// counts coins which have value 1
// we will need n coins of value 1
return n;
}
if (n > 9 && n < 25) {
// counts coins which have value 1 and 10
c = n / 10 + n % 10;
return c;
}
if (n > 24) {
// counts coins which have value 25
c = n / 25;
if (n % 25 < 10) {
// counts coins which have value 1 and 25
c = c + n % 25;
return c;
}
if (n % 25 > 9) {
// counts coins which have value 1, 10 and 25
c = c + (n % 25) / 10 + (n % 25) % 10;
return c;
}
}
}
// Driver Code
int main()
{
int n = 14;
cout << countCoins(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the minimum number
// of coins required
class GFG {
// Function to find the minimum number
// of coins required
static int countCoins(int n)
{
int c = 0;
if (n < 10) {
// counts coins which have value 1
// we will need n coins of value 1
return n;
}
if (n > 9 && n < 25) {
// counts coins which have value 1 and 10
c = n / 10 + n % 10;
return c;
}
if (n > 24) {
// counts coins which have value 25
c = n / 25;
if (n % 25 < 10) {
// counts coins which have value 1 and 25
c = c + n % 25;
return c;
}
if (n % 25 > 9) {
// counts coins which have value 1, 10 and 25
c = c + (n % 25) / 10 + (n % 25) % 10;
return c;
}
}
return c;
}
// Driver Code
public static void main(String[] args)
{
int n = 14;
System.out.println(countCoins(n));
}
}
Python 3
# Python3 program to find the minimum number
# of coins required
# Function to find the minimum number
# of coins required
def countCoins(n):
c = 0
if (n < 10):
# counts coins which have value 1
# we will need n coins of value 1
return n
if (n > 9 and n < 25):
# counts coins which have value 1 and 10
c = n // 10 + n % 10
return c
if (n > 24):
# counts coins which have value 25
c = n // 25
if (n % 25 < 10):
# counts coins which have value
# 1 and 25
c = c + n % 25
return c
if (n % 25 > 9):
# counts coins which have value
# 1, 10 and 25
c = (c + (n % 25) // 10 +
(n % 25) % 10)
return c
# Driver Code
n = 14
print(countCoins(n))
# This code is contributed by mohit kumar
C
// C# program to find the minimum number
// of coins required
using System;
class GFG
{
// Function to find the minimum number
// of coins required
static int countCoins(int n)
{
int c = 0;
if (n < 10)
{
// counts coins which have value 1
// we will need n coins of value 1
return n;
}
if (n > 9 && n < 25)
{
// counts coins which have value 1 and 10
c = n / 10 + n % 10;
return c;
}
if (n > 24)
{
// counts coins which have value 25
c = n / 25;
if (n % 25 < 10)
{
// counts coins which have value 1 and 25
c = c + n % 25;
return c;
}
if (n % 25 > 9)
{
// counts coins which have value 1, 10 and 25
c = c + (n % 25) / 10 + (n % 25) % 10;
return c;
}
}
return c;
}
// Driver Code
public static void Main()
{
int n = 14;
Console.WriteLine(countCoins(n));
}
}
// This code is contributed by Ryuga
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the minimum number
// of coins required
// Function to find the minimum number
// of coins required
function countCoins($n)
{
$c = 0;
if ($n < 10)
{
// counts coins which have value 1
// we will need n coins of value 1
return $n;
}
if ($n > 9 && $n < 25)
{
// counts coins which have value 1 and 10
$c = (int)($n / 10 + $n % 10);
return $c;
}
if ($n > 24)
{
// counts coins which have value 25
$c = (int)($n / 25);
if ($n % 25 < 10)
{
// counts coins which have value 1 and 25
$c = $c + $n % 25;
return $c;
}
if ($n % 25 > 9)
{
// counts coins which have value 1, 10 and 25
$c = $c + ($n % 25) / 10 + ($n % 25) % 10;
return $c;
}
}
return $c;
}
// Driver Code
$n = 14;
echo(countCoins($n));
// This code is contributed Code_Mech
java 描述语言
<script>
// JavaScript program to find the minimum number
// of coins required
// Function to find the minimum number
// of coins required
function countCoins( n)
{
var c = 0;
if (n < 10)
{
// counts coins which have value 1
// we will need n coins of value 1
return n;
}
if (n > 9 && n < 25)
{
// counts coins which have value 1 and 10
c = n / 10 + n % 10;
return Math.trunc(c);
}
if (n > 24)
{
// counts coins which have value 25
c = n / 25;
if (n % 25 < 10)
{
// counts coins which have value 1 and 25
c = c + n % 25;
return Math.trunc(c);
}
if (n % 25 > 9)
{
// counts coins which have value 1, 10 and 25
c = c + (n % 25) / 10 + (n % 25) % 10;
return Math.trunc(c);
}
}
}
var n = 14;
document.write(countCoins(n));
// This code is contributed by akshitsaxenaa09.
</script>
Output:
5
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