POJ 题目3181 Dollar Dayz（完全背包，技巧）

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4231		Accepted: 1646

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

Source

USACO 2006 January Silver

思路： http://blog.youkuaiyun.com/libin56842/article/details/9455979

<span style="color: rgb(0, 153, 0); font-family: 'ms shell dlg'; font-size: 14px; line-height: 28px; white-space: pre-wrap;">考虑边界状态:</span><br style="color: rgb(0, 153, 0); font-family: 'ms shell dlg'; font-size: 14px; line-height: 28px; white-space: pre-wrap; clear: both;" /><span style="color: rgb(0, 153, 0); font-family: 'ms shell dlg'; font-size: 14px; line-height: 28px; white-space: pre-wrap;">M = 1 或者 N = 1 只有一个划分 既: F(1,1) = 1</span><br style="color: rgb(0, 153, 0); font-family: 'ms shell dlg'; font-size: 14px; line-height: 28px; white-space: pre-wrap; clear: both;" /><span style="color: rgb(0, 153, 0); font-family: 'ms shell dlg'; font-size: 14px; line-height: 28px; white-space: pre-wrap;">M = N : 等于把M - 1 的划分数加 1 既: F(N,N) = F(N,N-1) + 1 </span><br style="color: rgb(0, 153, 0); font-family: 'ms shell dlg'; font-size: 14px; line-height: 28px; white-space: pre-wrap; clear: both;" /><span style="color: rgb(0, 153, 0); font-family: 'ms shell dlg'; font-size: 14px; line-height: 28px; white-space: pre-wrap;">M > N: 按理说,N 划分后的序列中最大数是不会超过 N 的,所以 F(N,M ) = F(N,N)</span><br style="color: rgb(0, 153, 0); font-family: 'ms shell dlg'; font-size: 14px; line-height: 28px; white-space: pre-wrap; clear: both;" /><span style="color: rgb(0, 153, 0); font-family: 'ms shell dlg'; font-size: 14px; line-height: 28px; white-space: pre-wrap;">M < N: 这个是最常见的, 他应该是序列中最大数为 M-1 的划分和 N-M 的划分之和, 比如F(6,4),上面例子第三行, 他应该等于对整数 3 的划分, 然后加上 2 的划分(6-4) 所以 F(N,M) = F(N, M-1) + F(N-M,M)</span>

ac代码

#include<stdio.h>
#include<string.h>
__int64 a[1010][110],b[1010][110],inf=1;
int main()
{
	int n,k,i;
	for(i=0;i<18;i++)
		inf*=10;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		int i,j;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(i=0;i<=k;i++)
			a[0][i]=1;
		for(j=1;j<=k;j++)
		{
			for(i=1;i<=n;i++)
			{
				if(i<j)
				{
					a[i][j]=a[i][j-1];
					b[i][j]=b[i][j-1];
				}
				else
				{
					a[i][j]=(a[i-j][j]+a[i][j-1])%inf;
					b[i][j]=b[i-j][j]+b[i][j-1]+(a[i-j][j]+a[i][j-1])/inf;
				}
			}
		}
		if(b[n][k])
			printf("%I64d",b[n][k]);
		printf("%I64d\n",a[n][k]);
	}
}