Uva 147 Dollars

最新推荐文章于 2018-08-16 16:07:24 发布

原创最新推荐文章于 2018-08-16 16:07:24 发布 · 698 阅读

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ACM/背包问题/完全背包专栏收录该内容

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本文探讨了如何解决完全背包问题，并通过优化输出格式来提高代码效率。主要关注于货币组合问题，实现了一个高效的算法来计算任意金额的组合方式数量。

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

简单的完全背包问题，要注意的是输出格式和数据范围

这样的完全背包问题求方案数，可以一次性都求出来，再进行直接读入与输出

#include<stdio.h>
#include<string.h>

long long f[30001];
long long c[11]={5,10,20,50,100,200,500,1000,2000,5000,10000};

int main(void)
{
	long long a,b,i;
	f[0]=1;
	for(int k=0;k<11;k++)
		for(i=c[k];i<=30000;i++)
		{
			if(f[i-c[k]])
				f[i]+=f[i-c[k]];
		}
	while(scanf("%lld.%lld",&a,&b),a||b)
	{
		long long v=a*100+b;	
		if(a<10)
			printf("  ");
		else
			if(a<100)
				printf(" ");
		printf("%lld.%02lld",a,b);
		long long tmp=f[v],count=16;
		while(tmp>=10)
		{
			tmp/=10;
			count--;
//			printf("tmp=%lld\n",tmp);
		}
		for(i=1;i<=count;i++)
		{
			printf(" ");
		}
		printf("%lld\n",f[v]);

	}
	return 0;
}