SCU 1009 Dollars

最新推荐文章于 2018-09-20 20:30:08 发布

Modiz

最新推荐文章于 2018-09-20 20:30:08 发布

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分类专栏： HUT训练

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本文介绍了一种通过暴力枚举的方法来解决新西兰货币不同面额组合的问题，旨在计算出特定金额下所有可能的组成方式数量。

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题解：据说是个背包问题；可惜我不是这么做的，我是暴力枚举！！！先把倍数乘上10

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 $50.00 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 5), followed by the number of ways in which that amount may be made up, right justified in a field of width 12.

Sample input

0.20
  2.00
0.00

Sample output

 0.20           4
 2.00         293

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
	double x;
	int i,j,k,l,n,m,o,p,q;
	while (~scanf("%lf",&x) && x)
	{
		x*=10;
		if (x<1) 
		{
			printf("%.2lf 1\n",x/10);
			continue;
		}
		int t=0;
		for (i=0;i<=x/500;i++)
		for (j=0;j<=(x-i*500)/200;j++)
		for (k=0;k<=(x-i*500-j*200)/100;k++)
		for (l=0;l<=(x-i*500-j*200-k*100)/50;l++)
		for (n=0;n<=(x-i*500-j*200-k*100-l*50)/20;n++)
		for (m=0;m<=(x-i*500-j*200-k*100-l*50-n*20)/10;m++)
		for (o=0;o<=(x-i*500-j*200-k*100-l*50-n*20-m*10)/5;o++)
		for (p=0;p<=(x-i*500-j*200-k*100-l*50-n*20-m*10-o*5)/2;p++)
		{    
			int y=x-i*500-j*200-k*100-l*50-n*20-m*10-o*5-p*2;
			t+=y+1;
		}		
		printf("%5.2lf%12d\n",x/10,t);		
	}
	return 0;
}