poj3181:藏坑的背包-优快云博客

本文链接：https://blog.youkuaiyun.com/csdnicewing/article/details/51605103

本文介绍了一种解决特定购物组合问题的方法，通过编程算法来计算在有限预算内购买不同价格商品的所有可能组合数量。该问题源于USACO比赛，采用动态规划技术实现，并分享了通过的代码示例。

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Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5926		Accepted: 2234

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

这道题如果背包做多了就有感觉。如果设dp[i][j]是看到第i种工具时能组成价格j的方案数，那么就有

dp[i][j]=dp[i-1][j-i]+dp[i-1][j-2*i]……一直到减到>=0

但是问题是这个问题的解，long long存不下，那么就得分成两块，一块前面的一块后面的，那个mod要足够大，因为最后一组数据是30位的数，mod弄小了会前面的存不下，缺位。

我还使用了循环的dp数组，要注意及时清空，不费时的。

下面贴代码：

Run ID	User	Problem	Result	Memory	Time	Language	Code Length	Submit Time
15596341	xxxx	3181	Accepted	196K	79MS	C++	821B	2016-06-07 17:07:29

#include
  
   
#include
   
    
#include
    
     
using namespace std;
long long n,k,mod;
int cur;
long long a[2][1010],b[2][1010];

int main()
{
	#ifdef ONLINE_JUDGE
	#else
	  freopen("ddayz.10.in","r",stdin);
	#endif
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	scanf("%lld %lld",&n,&k);
	a[0][0]=1;
	cur=0;mod=100000000000000000;
	for(int i=1;i<=k;i++)
	{
		cur=!cur;
		for(int j=0;j<=n;j++)
		{
			long long sum1=0,sum2=0;
			for(int m=0;m*i<=j;m++)
			  {
			  	sum1+=a[!cur][j-m*i];
			  	sum2+=b[!cur][j-m*i];
			  } 
			a[cur][j]=sum1;
			b[cur][j]=sum2;
			b[cur][j]+=a[cur][j]/mod;
			a[cur][j]=a[cur][j]%mod;
		} 
		memset(a[!cur],0,sizeof(a[!cur]));
		memset(b[!cur],0,sizeof(b[!cur]));
	}
	;
	if(b[cur][n]>0)
	  printf("%I64d%I64d",b[cur][n],a[cur][n]);
	else printf("%I64d",a[cur][n]);
	return 0;
}