【POJ - 3181】【Dollar Dayz】

题目：

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：

给我们两个数字，第一个是我们的金额，第二个代表从有1-m种金币，问由这些货币能组成现在第一个金额的组成方式。

解题思路：

完全背包+大数 因为数据范围会爆 long long 所以咱们开了二维的dp，dp[x][0]存放大数的前半段，dp[x][1]存放后半段，最后输出大树的前半段就好。

ac代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 2100
#define LIMIT_ULL 100000000000000000
using namespace std;
typedef long long ll;

ll dp[1005][2];
int main()
{
	int n,k;
	while(scanf("%d%d",&n,&k)==2)
	{
		memset(dp,0,sizeof(dp));
		dp[0][1]=1;
		for(int i=1;i<=k;i++)
			for(int j=i;j<=n;j++)
			{
				if(dp[j-i][1])
				{
					dp[j][0]+=dp[j-i][0];
					dp[j][1]+=dp[j-i][1];
					dp[j][0]+=dp[j][1]/LIMIT_ULL;
					dp[j][1]=dp[j][1]%LIMIT_ULL;
				}
			}
		if(dp[n][0])
		{
			cout<<dp[n][0];
		}
		cout<<dp[n][1]<<endl;
	}
	return 0;
}