Lucas theorem

Ans=c (n,0) +c (n,1) +......+c (n,k)

=c (n/2333,0) *c (n%2333,0) +c (n/2333,0) *c (n%2333,1) +......+c (n/2333,0) *c (n%2333,2332) +c (/2333,1) *C (n%2333,0) +......+C (n/) 2333,k/2333) * (n%2333,k%2333)

=∑c (N/2333,J) *sum[n%2333][2332] (0<=j<k/2333) +c (n/2333,k/2333) *sum[n%2333][k%2333]

Cal (N,k) =cal (n/2333,k/2333-1) *sum[n%2333][2332]+lucas (n/2333,k/2333) *sum[n%2333][k%2333]

Attention to prefix and processing

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
# include<algorithm>
#include <iostream>
#define MOD 2333
using namespace std;
int T;
Long long c[mod][mod],sum[mod][mod];
Long Long Lucas (long long X,long long y)
{
if (x<y | | | y<0) return 0;
if (x<mod && y<mod) return c[x][y];
Return Lucas (X/mod,y/mod) *c[x%mod][y%mod]%mod;
}
Long Long cal (Long long N,long long k)
{
if (k<0) return 0;
Return (Cal (n/mod,k/mod-1) *sum[n%mod][mod-1]+lucas (n/mod,k/mod) *sum[n%mod][k%mod])%mod;
}
int main ()
{
c[0][0]=1;sum[0][0]=1;
for (int i=1;i<mod;i++) sum[0][i]=1;
for (int i=1;i<mod;i++)
{
c[i][0]=sum[i][0]=1;
for (int j=1;j<=i;j++) c[i][j]= (c[i-1][j-1]+c[i-1][j])%mod;
for (int j=1;j<mod;j++) sum[i][j]= (sum[i][j-1]+c[i][j])%mod;
}
scanf ("%d", &t);
while (t--)
{
long long n,k;
scanf ("%lld%lld", &n,&k);
printf ("%lld\n", Cal (N,k));
return 0;
}