[2019.2.24]BZOJ4591 [Shoi2015]超能粒子炮·改

以下除法一律为整除。

求\(\sum_{i=0}^kC_n^i\ mod\ p,p=2333\)

设\(f(i,j)=\sum_{k=0}^jC_i^k\)

\(f(n,k)=\sum_{i=0}^kC_n^i\)

当\(k\le p,n\le p\),预处理\(0\le i\le p,0\le j\le i\)的\(C_i^j\)及其前缀和。

否则

\(f(n,k)\)

\(=\sum_{i=0}^kC_{n\ mod\ p}^{i\ mod \ p}\times C_{n/p}^{i/p}\)

\(=\sum_{b=0}^{k/p-1}\sum_{i=0}^{p-1}(C_{n\ mod\ p}^i\times C_{n/p}^{b})+C_{n/p}^{k/p}\times(\sum_{i=0}^{k\ mod\ p}C_{n\ mod\ p}^i)\)

\(=\sum_{b=0}^{k/p-1}C_{n/p}^{b}\times\sum_{i=0}^{p-1}C_{n\ mod\ p}^i+C_{n/p}^{k/p}\times(\sum_{i=0}^{k\ mod\ p}C_{n\ mod\ p}^i)\)

\(=f(n/p,k/p-1)\times \sum_{i=0}^{p-1}C_{n\ mod\ p}^i+C_{n/p}^{k/p}\times f(n\ mod\ p,k\ mod\ p)\)

\(f(n/p,k/p-1)\)递归,\(\sum_{i=0}^{p-1}C_{n\ mod\ p}^i\)预处理过了,\(C_{n/p}^{k/p}\)直接lucas定理,\(f(n\ mod\ p,k\ mod\ p)\)也预处理过了。

code:

#include<bits/stdc++.h>
using namespace std;
const long long p=2333;
int T,C[p+10][p+10],sum[p+10][p+10];
long long n,k;
int Lucas(long long x,long long y){//C_x^y
	if(y>x)return 0;
	if(x<=p&&y<=p)return C[x][y];
	return Lucas(x/p,y/p)*C[x%p][y%p]%p;
}
int F(long long x,long long y){
	if(y>x)y=x;
	if(!y)return 1;
	if(y<0)return 0;
	if(x<=p&&y<=p)return sum[x][y];
	return (F(x/p,y/p-1)*sum[x%p][p-1]%p+Lucas(x/p,y/p)*sum[x%p][y%p]%p)%p;
}
int main(){
	C[0][0]=sum[0][0]=1;
	for(int i=1;i<=p;++i)for(int j=0;j<=i;++j)C[i][j]=(C[i-1][j]+(j?C[i-1][j-1]:0))%p;
	for(int i=0;i<=p;++i)for(int j=0;j<=p;++j)sum[i][j]=((j?sum[i][j-1]:0)+C[i][j])%p;
	scanf("%d",&T);
	while(T--)scanf("%lld%lld",&n,&k),printf("%d\n",F(n,k));
	return 0;
}