Dollar Dayz （大数dp fuck！不是多组数据！！）

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.

#include<bits/stdc++.h>
using namespace std;
const long long inf=1e15;
long long num[100];
long long dp[1010][110][10];

void add(int a,int b,int c,int d)  //每位存15位数
{
    long long mx=0;
    for(int i=0; i<3; i++)
    {
        long long z=dp[a][b][i]+dp[c][d][i]+mx;
        dp[a][b][i]=z%inf;
        mx=z/inf;
    }
}

void print(int a,int b)//
{
    int t=0;
    for(int i=0; i<3; i++)
    {
        num[i]=dp[a][b][i];
        //printf("%d %d %d %dn",a,b,i,dp[a][b][i]);
        if(dp[a][b][i]!=0)
            t=i;
    }
    for(int i=t; i>=0; i--)
    {
        if(i!=t)
            printf("%015lld",num[i]);
        else
            printf("%lld",num[i]);
    }
    printf("n");
}

int main()
{


    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        memset(dp,sizeof(dp));
        //printf("*****n");
        for(int i=1; i<=n+1; i++)
        {
            dp[i][1][0]=1;
            int aaa=min(i,k);
            for(int j=2; j<=aaa; j++)
            {
                for(int k=1; k<=j; k++)
                {
                    if(i-k<k)
                        add(i,j,i-k,i-k);    //dp[i][j] = dp[i][j] + dp[i-k][i-k];
                    else
                        add(i,k);      // dp[i][j] = dp[i][j] + dp[i-k][k];
                }
            }

        }

        int t=0;
        for(int i=0; i<3; i++)
        {
            num[i]=dp[n+1][k][i];
            //printf("%d %d %d %dn",dp[a][b][i]);
            if(dp[n+1][k][i]!=0)
                t=i;
        }
        for(int i=t; i>=0; i--)
        {
            if(i!=t)
                printf("%015lld",num[i]);//补零
            else
                printf("%lld",num[i]);
        }
        printf("n");
    }
}