Poj 3181 Dollar Dayz

最新推荐文章于 2022-04-18 19:42:48 发布

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本文探讨了如何通过大数运算解决完全背包问题，重点在于利用基本的算法技巧来处理大数值的加法操作，从而实现有效计算不同组合方式的解决方案总数。实例分析了FJ在Cow Store购买工具的场景，展示了算法应用的实际效果。

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Description

Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $1, $2, and $3. Farmer John has exactly $5 to spend. He can buy 5 tools at $1 each or 1 tool at $3 and an additional 1 tool at $2. Of course, there are other combinations for a total of 5 different ways FJ can spend all his money on tools. Here they are:

1 @ US$3 + 1 @ US$2

1 @ US$3 + 2 @ US$1

1 @ US$2 + 3 @ US$1

2 @ US$2 + 1 @ US$1

5 @ US$1

Write a program than will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of $1..$K (1 <= K <= 100).

Input

A single line with two space-separated integers: N and K.

Output

A single line with a single integer that is the number of unique ways FJ can spend his money.

Sample Input

5 3

Sample Output

Source

USACO 2006 January Silver

完全背包问题，数据比较大，需要运用简单的大数运算

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

#define MAXN 100
int f[1001][MAXN];

void fun(int a[],int b[])          //大数相加
{
	int i,tmp,m=0;
	for(i=0;i<MAXN;i++)
	{
		tmp=a[i]+b[i]+m;
		a[i]=tmp%10;
		m=tmp/10;
	}
}

int main(void)
{
	int N,K;
//	freopen("d:\\in.txt","r",stdin);
	while(scanf("%d%d",&N,&K)==2)
	{
		int i,j;
//		memset(f,0,(N+1)*sizeof(f[0]));
		memset(f,0,sizeof(f));
		f[0][0]=1;
		for(i=1;i<=K;i++)
			for(j=i;j<=N;j++)
			{
					fun(f[j],f[j-i]);
			}
		i=MAXN;
		while(!f[N][i])  i--;
		while(i>=0)
			printf("%d",f[N][i--]);
		printf("\n");
	}
	return 0;
}