I - Dollar Dayz

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

The general meaning of the topic is ： use n Money to buy value 1~k The items , Items can be purchased repeatedly , How many combinations are there at most

The recursive formula 1.i>=j dp[I][j]=dp[I][j-1]+dp[I-j][j]

               2.i<j dp[I][j]=dp[I][I]

Because the result is relatively large So use two ll The array stores the results

One problem that hasn't been solved is the final output If the low order is lower than 17 position , The high position is not equal to 0, This output is obviously wrong , Please answer me

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define ll long long
ll a[10009][108],b[10009][106];
ll inf;
int main()
{
    int n,m,i,j;
    inf=1;
    for(i=1;i<=18;i++)
    {
        inf=inf*10;
    }
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=1;i<=m+1;i++)
        {
            a[0][i]=1;
        }
        for(i=1;i<=n+1;i++)
        {
            a[i][1]=1;
        }
        for(i=1;i<=m+1;i++)
        {
            a[1][i]=1;
        }
        for(i=2;i<=n;i++)
        {
            for(j=2;j<=m;j++)
            {
                if(i<j)
                {
                    a[i][j]=a[i][i];
                    b[i][j]=b[i][i];
                }
                else
                {
                    a[i][j]=(a[i][j-1]+a[i-j][j])%inf;
                    b[i][j]=b[i][j-1]+b[i-j][j]+(a[i][j-1]+a[i-j][j])/inf;
                }//printf("%d ",a[i][j]);
            }
        }
        if(b[n][m]!=0)
            printf("%lld",b[n][m]);
        printf("%lld\n",a[n][m]);
    }
}