Dollar Dayz POJ - 3181

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

题解：
　　完全背包，dp[j]表示凑出j的方案数，dp[j]=dp[j]+dp[j-i],方案数等于不用i的方案数加上用了i的方案数。
　　只是这个题目会报ll所以要开两个数组分别存前18位和后18位。
代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#define ll long long
#define inf 1000000000000000000
using namespace std;
ll dp[2000],f[2000];
int main()
{
    int n,k;scanf("%d%d",&n,&k);
    dp[0]=1;
    for(int i=1;i<=k;i++)
        for(int j=1;j<=n;j++){
            if(j<i) continue;
            f[j]=f[j]+f[j-i]+(dp[j]+dp[j-i])/inf;
            dp[j]=(dp[j]+dp[j-i])%inf;
        }
    if(!f[n]) printf("%lld",dp[n]); 
    else printf("%lld%lld",f[n],dp[n]);
    return 0;
}