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).

A single line with two space-separated integers: N and K.

A single line with a single integer that is the number of unique ways FJ can spend his money.

5 3

5

#include <iostream>

#include <queue>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <stack>

#include <vector>

#include <algorithm>

using namespace std;

#define N 2100

#define met(a,b) (memset(a,b,sizeof(a)))

typedef long long LL;

LL dp[N][];

int main()

{

    int n, m;

    while(scanf("%d%d", &n, &m)!=EOF)

    {

        int i, j;

        met(dp, );

        dp[][] = ;

        for(i=; i<=m; i++)

        for(j=i; j<=n; j++)

        {

            dp[j][] += dp[j-i][];

            dp[j][] += dp[j-i][];

            dp[j][] += dp[j][]/;

            dp[j][] %= ;

        }

        if(dp[n][]==)

            printf("%I64d\n", dp[n][]);

        else

        {

            printf("%I64d%015I64d\n", dp[n][], dp[n][]);

        }

    }

    return ;

}