POJ 3181 Dollar Dayz(高精度完全背包)

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5342		Accepted: 2025

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

5

题意：有n元钱，商品的价格在1~k元（每种价格的商品数量无限），用n元去买这些商品，最多有多少种选择。

题解：很明显的完全背包啦，不过当n为1000，k=100时，结果为32整数，超过了long long。 有两种解决方法：

1.直接在dp方程相加转化时模拟大数加法，个人就是这么写的

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[1010][100];

int add(int a[],int b[])
{
	int k;
	for(k=0;k<100;++k)
	{
		b[k]+=a[k];
		if(b[k]>9)
		{
			b[k+1]++;
			b[k]-=10;
		}
	}
	return *b;
}

int main()
{
	int n,i,j,k;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		dp[0][0]=1;
		for(i=1;i<=k;++i)
		{
			for(j=i;j<=n;++j)
				*dp[j]=add(dp[j-i],dp[j]);
		}
		for(i=100;i>0;--i)
			if(dp[n][i])
				break;
		for(;i>=0;i--)
			printf("%d",dp[n][i]);
		printf("\n");
	}
	return 0;
}

2，这种方法是膜大神得来的，用两个long long数组拼接来表示超过long long的数据，long long为19位整数，注意输出和进位。

代码如下：

#include<cstdio>
#include<cstring>
#define ll __int64
ll INF=1000000000000000000;
ll a[1010];//高位 
ll b[1010];//低位 
int main()
{
	int n,k,i,j;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		b[0]=1;
		for(i=1;i<=k;++i)
		{
			for(j=i;j<=n;++j)
			{
				a[j]=a[j]+a[j-i]+(b[j]+b[j-i])/INF;
				b[j]=(b[j]+b[j-i])%INF;
			}
		}
		if(a[n]==0)
			printf("%I64d\n",b[n]);
		else
			printf("%I64d%018I64d\n",a[n],b[n]);
	}
	return 0;
}