BZOJ 3992: [SDOI2015]序列统计

NTT模板题解析

最新推荐文章于 2019-07-04 22:58:56 发布

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最新推荐文章于 2019-07-04 22:58:56 发布

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原文链接：http://www.cnblogs.com/silenty/p/10319213.html

NTT模板题

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring> 
#define ll long long
using namespace std;
const int N=20000;
const int mod=1004535809;
int L,M,n,m,x,s,w,len,lenf,X,vis[N],rev[N];
ll h[N],a[N],b[N],q[N];
ll pow(ll a,ll b,ll mod){
	a%=mod;
	ll ans=1;
	while (b){
		if (b&1) ans=ans*a%mod;
		a=a*a%mod;
		b=b>>1;
	}
	return ans;
}
int calc_w(int x){
	if (x==2) return 1;
	for (int i=2; i; i++){
		int flag=1;
		for (int j=2; j*j<x; j++)
			if (pow(i,(x-1)/j,x)==1){
				flag=0;
				break;
			}
		if (flag) return i;
	}
}
void NTT(ll *a,int n,int opt){
    for (int i=0; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
	for (int i=1; i<n; i<<=1){
		ll wn=pow(3,(mod-1)/(i<<1),mod);
		for (int j=0; j<n; j+=(i<<1)){
			ll w=1;
			for (int k=0; k<i; k++,w=w*wn%mod){
				ll x=a[j+k],y=w*a[j+k+i]%mod;
				a[j+k]=(x+y)%mod,a[j+k+i]=(x-y+mod)%mod;
			}
		}
	}
	if (opt==-1) reverse(a+1,a+n);
}
void mul(ll h[N],ll q[N]){
	for (int i=0; i<len; i++) a[i]=h[i]%mod;
	for (int i=0; i<len; i++) b[i]=q[i]%mod;
	NTT(a,len,1); 
	NTT(b,len,1);		
	for (int i=0; i<len; i++) a[i]=a[i]*b[i]%mod;
	NTT(a,len,-1);
	ll inv=pow(len,mod-2,mod);
	for (int i=0; i<len; i++) a[i]=a[i]*inv%mod;
	for (int i=0; i<lenf; i++) h[i]=(a[i]+a[i+lenf])%mod;
}
void solve(int x){
	h[0]=1;
	while (x){
		if (x&1) mul(h,q);
		x>>=1;
		mul(q,q);
	}
}
int main(){
	scanf("%d%d%d%d",&n,&m,&X,&s);
	w=calc_w(m);
	for (int i=1; i<=s; i++){
		int x;
		scanf("%d",&x);
		vis[x]=1;
	}
	int pos=-1;
	for (int i=0,j=1; i<m-1; i++,j=j*w%m){
		if (vis[j]) q[i]=1;
		if (j==X) pos=i;
	}
	lenf=m-1;
	M=lenf*2,len=1,L=0;
	while (len<=M) len<<=1,L++;
	for (int i=0; i<len; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
	solve(n);
	if (pos!=-1) printf("%lld\n",h[pos]);
	else printf("0\n");
	return 0;
}

转载于:https://www.cnblogs.com/silenty/p/10319213.html