loj2002 「SDOI2017」序列计数

水题

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long ll;
int n, m, p, cnt[105], pri[2000005], ppp, ans=0;
const int mod=20170408;
bool isp[20000005];
struct Matrix{
	int num[105][105];
	Matrix operator*(const Matrix &x)const{
		Matrix re;
		for(int i=0; i<p; i++)
			for(int j=0; j<p; j++){
				re.num[i][j] = 0;
				for(int k=0; k<p; k++)
					re.num[i][j] = (re.num[i][j] + (ll)num[i][k]*x.num[k][j]) % mod;
			}
		return re;
	}
}yua, dan, zhu;
Matrix ksm(Matrix a, int b){
	Matrix re=dan;
	while(b){
		if(b&1)	re = re * a;
		a = a * a;
		b >>= 1;
	}
	return re;
}
void shai(){
	memset(isp, true, sizeof(isp));
	isp[0] = isp[1] = false;
	for(int i=2; i<=m; i++){
		if(isp[i])	pri[++ppp] = i;
		for(int j=1; j<=ppp && (ll)i*pri[j]<=m; j++){
			isp[i*pri[j]] = false;
			if(i%pri[j]==0)	break;
		}
	}
}
int main(){
	cin>>n>>m>>p;
	shai();
	for(int i=1; i<=m; i++)
		cnt[i%p]++;
	yua.num[0][0] = 1;
	for(int i=0; i<p; i++){
		dan.num[i][i] = 1;
		for(int j=0; j<p; j++)
			zhu.num[i][j] = cnt[((i-j)%p+p)%p];
	}
	ans = (yua*ksm(zhu, n)).num[0][0];
	for(int i=1; i<=ppp; i++)
		cnt[pri[i]%p]--;
	for(int i=0; i<p; i++){
		dan.num[i][i] = 1;
		for(int j=0; j<p; j++)
			zhu.num[i][j] = cnt[((i-j)%p+p)%p];
	}
	ans = ((ans - (yua*ksm(zhu, n)).num[0][0])%mod + mod) % mod;
	cout<<ans<<endl;
	return 0;
}