According to Lucas ' theorem, we have

Ans (n,k) =ans (⌊np⌋,⌊kp⌋−1) ∗∑i=0p−1 (N%PI) + (⌊n/p⌋⌊k/p⌋) ∑i=0k%p (n%pi) ans (n,k) =ans (\lfloor\frac n p \rfloor,\lfloor \ Frac k p\rfloor-1) *\sum_{i=0}^{p-1}\binom{n\%p}i+\binom{\lfloor N/p\rfloor}{\lfloor k/p\rfloor}\sum_{i=0}^{k\%p}\ Binom{n\%p}i

One of the ans-ans-term recursive computations, combined number with Lucas violence calculation, and type O (p2) o (p^2) pretreatment.

Complexity O (P2+TLOGNLOGP) o (P^2+t\log n\log p).

#include <cstdio> #include <algorithm> using namespace std;
#define LL Long Long const int p=2333;
int pow (int base,int k) {int ret=1;
for (; k;k>>=1,base=base*base%p) if (k&1) ret=ret*base%p;
return ret;
} int fac[p+10],inv[p+10],sum[p+10][p+10];
int C (LL n,ll k) {if (k>n) return 0;
if (n<p&&k<p) return fac[n]*inv[k]%p*inv[n-k]%p;
return C (n%p,k%p) *c (n/p,k/p)%p;
} int Solve (LL n,ll k) {int ret=0,tem=0;
if (n<p&&k<p) return sum[n][k];
Return (Sum[n%p][p-1]*solve (n/p,k/p-1)%p+sum[n%p][k%p]*c (n/p,k/p)%p)%p;
} int main () {LL n,k;
Inv[0]=fac[0]=1;
for (int i=1;i<p;i++) fac[i]=fac[i-1]*i%p;
Inv[p-1]=pow (fac[p-1],p-2);
for (int i=p-2;i;i--) inv[i]=inv[i+1]* (i+1)%p;
for (int i=0;i<p;i++) {sum[i][0]=1;
for (int j=1;j<p;j++) sum[i][j]= (Sum[i][j-1]+c (i,j))%p;
} int T;
scanf ("%d", &t); while (t--) {scanf ("%lld%lld",&N,&K);
printf ("%d\n", Solve (n,k)); }
}