BZOJ 3992 Sdoi2015 sequential statistics fast number theory transformation, bzoj3992

BZOJ 3992 Sdoi2015 sequential statistics fast number theory transformation, bzoj3992

Given n (n <= 10 ^ 9), prime number m (3 <= m <= 8000), 1 <= x = m, and a [0, the number of series with the length of n in the S set in the m-1] interval satisfies that each element belongs to the S set and the product mod m of all elements is equal to x.

Evaluate the original root, obtain the index for each element in the S set, and then generate the f (x) function)

The answer is (f (x) ^ n (mod x ^ m-1), mod 1004535809)

NTT just needs to use a polynomial to quickly create a power.

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define M 16400#define MOD 1004535809#define INF 0x3f3f3f3f#define G 3using namespace std;int n,m,d,X,S;int ind[M];long long Quick_Power(long long x,int y,int p){long long re=1;while(y){if(y&1) (re*=x)%=p;(x*=x)%=p; y>>=1;}return re;}void NTT(int a[],int n,int type){static int temp[M];int i;if(n==1) return ;for(i=0;i<n;i+=2)temp[i>>1]=a[i],temp[i+n>>1]=a[i+1];memcpy(a,temp,sizeof(a[0])*n);int *l=a,*r=a+(n>>1);NTT(l,n>>1,type);NTT(r,n>>1,type);long long w=Quick_Power(G,(long long)type*(MOD-1)/n%(MOD-1),MOD),wn=1;for(i=0;i<n>>1;i++,(wn*=w)%=MOD)temp[i]=(l[i]+wn*r[i])%MOD,temp[i+(n>>1)]=(l[i]-wn*r[i]%MOD+MOD)%MOD;memcpy(a,temp,sizeof(a[0])*n);}struct GF{int a[M];GF() {}GF(bool){memset(a,0,sizeof a);a[0]=1;}int& operator [] (int x){return a[x];}GF& operator *= (const GF &f){static int b[M];int i;memcpy(b,f.a,sizeof b);NTT(a,d,1);NTT(b,d,1);for(i=0;i<d;i++)a[i]=(long long)a[i]*b[i]%MOD;NTT(a,d,MOD-2);for(i=m-1;i<=m-2<<1;i++)(a[i-(m-1)]+=a[i])%=MOD,a[i]=0;long long inv=Quick_Power(d,MOD-2,MOD);for(i=0;i<=m-2;i++)a[i]=a[i]*inv%MOD;return *this;}}a;int Get_Primitive_Root(){static int stack[M],top;int i,j,temp=m-1;for(i=2;i<=temp;i++)if(temp%i==0){stack[++top]=i;while(temp%i==0)temp/=i;}for(i=2;;i++){for(j=1;j<=top;j++)if(Quick_Power(i,(m-1)/stack[j],m)==1)break;if(j==top+1)return i;}}GF Quick_Power(GF &x,int y){GF re(true);while(y){if(y&1) re*=x;x*=x; y>>=1;}return re;}int main(){int i,x;cin>>n>>m>>X>>S;for(d=1;d<=m+m;d<<=1);int g=Get_Primitive_Root();for(i=0,x=1;i<m-1;i++,(x*=g)%=m)ind[x]=i;for(i=1;i<=S;i++){scanf("%d",&x);if(!x) continue;a[ind[x]]=1;}GF ans=Quick_Power(a,n);cout<<ans[ind[X]]<<endl;return 0;}