Dollar Dayz POJ

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;
}