UVA 147 Dollars

最新推荐文章于 2021-02-01 21:00:38 发布

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最新推荐文章于 2021-02-01 21:00:38 发布

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原文链接：http://www.cnblogs.com/yspworld/p/4320787.html

本文介绍了一道关于计算新西兰币不同面额组合数量的编程题。输入为一系列不超过300.00的新西兰币金额，输出为该金额可由多少种不同面额组合而成。文章提供了一个C++实现的示例代码。

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Dollars

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c coins. Write a program that will determine, for any given amount, in how many ways that amount may be made up. Changing the order of listing does not increase the count. Thus 20c may be made up in 4 ways: 1 20c, 2 10c, 10c+2 5c, and 4 5c.

Input

Input will consist of a series of real numbers no greater than $300.00 each on a separate line. Each amount will be valid, that is will be a multiple of 5c. The file will be terminated by a line containing zero (0.00).

Output

Output will consist of a line for each of the amounts in the input, each line consisting of the amount of money (with two decimal places and right justified in a field of width 6), followed by the number of ways in which that amount may be made up, right justified in a field of width 17.

Sample input

0.20
2.00
0.00

Sample output

  0.20                4
  2.00              293

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <algorithm>

using namespace std;
long long a[6001]; //记录每种钱的组合数目
int b[]={1,2,4,10,20,40,100,200,400,1000,2000};

int main()
{
	for(int i=0; i<=6000; i++)
		a[i]=1; //仅由5分构成的话
	for(int i=1; i<11; i++) //从下标1开始，0下标已被跳过
	{
		for(int j=b[i]; j<=6000; j++) //枚举可使用第i类货币每一种可能
		{
			a[j]=a[j]+a[j-b[i]]; //当前的种类数+前i-1种货币构成j-b[i]的方法数
		}
	}
	double dd;
	while(scanf("%lf", &dd)!=EOF)
	{
		if(dd==0.00)
			break;
		int n=(int)(dd*20.0);
		printf("%6.2lf%17lld\n", dd, a[n] );
	}
	return 0;
}

转载于:https://www.cnblogs.com/yspworld/p/4320787.html