由Lucas定理C(n,k)=C(n/2333,k/2333)*C(n%2333,k%2333)%2333
则ans=ΣC(n,i),(i<=k)
=C(n/2333,0)*C(n%2333,0)+C(n/2333,1)*C(n%2333,1)+...+C(n/2333,2332)
+C(n/2333,0)*C(n%2333,0)+C(n/2333,1)*C(n%2333,1)+...+C(n/2333,2332)
=∑C(n/2333,j)*sum[n%2333][2332]+C(n/2333,k/2333)*sum[n%2333][k%2333],(0<=j<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]
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #define int long long using namespace std; const int p=2333; int T,c[p+10][p+10],sum[p+10][p+10]; int Lucas(int a,int b) { if(a<0||b<0) return 0; if(a<p&&b<p) return c[a][b]; return Lucas(a/p,b/p)*c[a%p][b%p]%p; } int cal(int n,int k) { if(k<0) return 0; return (cal(n/p,k/p-1)*sum[n%p][p-1]+Lucas(n/p,k/p)*sum[n%p][k%p])%p; } void pre() { c[0][0]=1;sum[0][0]=1; for(int i=1;i<p;i++) c[i][0]=1,sum[i][0]=1,sum[0][i]=1; for(int i=1;i<p;i++) for(int j=1;j<=i;j++) c[i][j]=(c[i-1][j-1]+c[i-1][j])%p; for(int i=1;i<p;i++) for(int j=1;j<p;j++) sum[i][j]=sum[i][j-1]+c[i][j]; } signed main() { pre(); scanf("%lld",&T); while(T--) { int n,k; scanf("%lld%lld",&n,&k); printf("%lld\n",cal(n,k)); } return 0; }
时间: 2024-11-12 02:30:38