Report problem solving: luogu P2822

Topic link: P2822 combination of several problems 
it's a two-dimensional prefix and water, we only need a combination of the number of recursive ** (Pascal's Triangle) ** like, is this:
$$ C_n ^ = C_ {the n-m-1} ^ m + C_ {n-1 } ^ {m-1} $$
like, film edge to take recursive side, and the number of maintenance can be two-dimensional prefix.
(I will not tell you I was wrong cycle boundary $ WA $ the $ n $ times
$ Code $:

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
long long f[2005][2005],sum[2005][2005];
long long ch[2005][2005],cs[2005][2005];
int t,n,m,k;
int main()
{
    //freopen("data.in","r",stdin);
    //freopen("baoli.out","w",stdout);
    scanf("%d%d",&t,&k);
    for(int i=0;i<=2001;i++)
    {
        f[i][0]=1;
        ch[i][0]=0;
        ch[0][i]=0;
        cs[0][i]=0;
        cs[0][i]=0;
        sum[i][0]=0;
        sum[0][i]=0;
    }
    for(int i=1;i<=2001;i++)
    {
        for(int j=1;j<=i;j++)
        {
            f[i][j]=(f[i-1][j]+f[i-1][j-1])%k;
        }
    }
    for(int i=1;i<=2001;i++)
    {
        for(int j=1;j<=i;j++)
        {
            if(f[i][j]==0) f[i][j]=1;
            else f[i][j]=0;
        }
    }
    for(int i=1;i<=2001;i++)
    {
        for(int j=1;j<=i;j++)
        {
            ch[i][j]=ch[i][j-1]+f[i][j];
        }
    }
    for(int i=1;i<=2001;i++)
    {
        for(int j=i;j<=2001;j++)
        {
            cs[j][i]=cs[j-1][i]+f[j][i];
        }
    }
    for(int i=1;i<=2001;i++)
    {
        for(int j=1;j<=i;j++)
        {
            sum[i][j]=sum[i-1][j-1]+cs[i][j]+ch[i][j]-f[i][j];
        }
    }
    while(t--)
    {
        scanf("%d%d",&n,&m);
        if(n==0||m==0) cout<<0<<endl;
        else printf("%lld\n",sum[n][min(n,m)]);
    }
    return 0;
}
/*1 12
1962 1183*/

I look at the code stamp