POJ 3181 Dollar Dayz

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:8242		Accepted: 3082

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

dp[i][j]表示花j元从前i种工具中购买的方法数
则有两种情况，
1.第i种工具的花费大于j则不能买第i种，dp[i][j]=dp[i-1][j]。
2.可以买第i种，则dp[i][j]=dp[i-1][j]+dp[i][j-1]。
注意因为此数较大，会超过 long long 的范围，因此用大数运算
而开三维数组可能会导致内存过大，所以此处dp使用滚动数组。
代码：
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<cstdio>
int dp[2][1002][40];
using namespace std;
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        int i,j,k;
        memset(dp,0,sizeof(dp));
        for(i=0; i<2; i++)
            dp[i][0][0]=1;
        for(i=1; i<=m; i++)
        {
            for(j=1; j<=n; j++)
            {
                if(j>=i)
                {
                    int x=0;
                    for(k=0;k<40;k++)
                    {
                        dp[i%2][j][k]=dp[(i-1)%2][j][k]+dp[i%2][j-i][k]+x;
                        x=dp[i%2][j][k]/10;
                        dp[i%2][j][k]%=10;
                    }
                }
                else
                {
                    for(k=0;k<40;k++)
                        dp[i%2][j][k]=dp[(i-1)%2][j][k];
                }
            }
        }
        int L=39;
        while(dp[m%2][n][L]==0)
            L--;
        for(i=L;i>=0;i--)
            printf("%d",dp[m%2][n][i]);
    }
    return 0;
}