COGS 2294. [HZOI 2015] 释迦

额，其实就是裸的三模数NTT，上一篇已经说过了

哦，还有一个就是对乘起来炸long long的数取模，用long double之类的搞一下就好，精度什么的，，（看出题人心情？？）

 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #include<cstring>
 #define LL long long
 #define N 300005
 using namespace std;
 inline int ra()
 {
     int x=; char ch=getchar();
     while (ch<'' || ch>'') ch=getchar();
     while (ch>='' && ch<='') {x=x*+ch-''; ch=getchar();}
     return x;
 }

 const int mod=;
 const int M[]={,,};
 const int G[]={,,};
 const LL _M=(LL)M[]*M[];

 LL ksm(LL a, int b, int P)
 {
     LL sum=;
     for (;b;b>>=,a=a*a%P)
         if (b&) sum=sum*a%P;
     return sum;
 }
 /*LL mul(LL a, LL b, LL p)
 {
     a%=p; b%=p;
     return ((a*b-(LL)((LL)((long double)a/p*b+1e-3)*p))%p+p)%p;
 }
 const int m1=M[0],m2=M[1],m3=M[2];
 const int inv1=ksm(m1%m2,m2-2,m2),inv2=ksm(m2%m1,m1-2,m1),inv12=ksm(_M%m3,m3-2,m3);
 int CRT(int a1, int a2, int a3)
 {
     LL A=(mul((LL)a1*m2%_M,inv2,_M)+mul((LL)a2*m1%_M,inv1,_M))%_M;
     LL k=((LL)a3+m3-A%m3)*inv12%m3;
     return (k*(_M%mod)+A)%mod;
 }*/

 inline LL mul(LL a,LL b,LL p){
   a%=p; b%=p;
   return ((a*b-(LL)((LL)((long double)a/p*b+1e-)*p))%p+p)%p;
 }

 const int m1=M[],m2=M[],m3=M[];
 const int inv1=ksm(m1%m2,m2-,m2),inv2=ksm(m2%m1,m1-,m1),inv12=ksm(_M%m3,m3-,m3);
 inline int CRT(int a1,int a2,int a3){
   LL A=(mul((LL)a1*m2%_M,inv2,_M)+mul((LL)a2*m1%_M,inv1,_M))%_M;
   LL k=((LL)a3+m3-A%m3)*inv12%m3;
   return (k*(_M%mod)+A)%mod;
 }
 int rev[N];
 struct NTT{
     int num,P,G;
     int w[][N];
     void pre(int _P, int _G, int n)
     {
         num=n; P=_P; G=_G;
         int g=ksm(G,(P-)/num,P);
         w[][]=; for (int i=; i<n; i++) w[][i]=(LL)w[][i-]*g%P;
         w[][]=; for (int i=; i<n; i++) w[][i]=w[][n-i];
     }
     void FFT(int *a, int n, int f)
     {
         for (int i=; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
         for (int i=; i<n; i<<=)
             for (int j=; j<n; j+=(i<<))
                 for (int k=; k<i; k++)
                 {
                     int x=a[j+k],y=(LL)w[f][num/(i<<)*k]*a[i+j+k]%P;
                     a[j+k]=(x+y)%P; a[i+j+k]=(x-y+P)%P;
                 }
         if (!f) for (int i=,inv=ksm(n,P-,P); i<n; i++) a[i]=(LL)a[i]*inv%P;
     }
 }ntt[];

 int n,m;
 int ans[][N];
 int a[N],b[N],c[N],d[N];

 int main()
 {
       freopen("annona_squamosa.in","r",stdin); freopen("annona_squamosa.out","w",stdout);
     n=ra();
     for (int i=; i<n; i++) a[i]=ra();
     for (int i=; i<n; i++) b[i]=ra();
     for (m=; m<n<<; m<<=);
     int L=,x=m>>; while (!(x&)) x>>=,L++;
     for (int i=; i<m; i++) rev[i]=(rev[i>>]>>)|((i&)<<L);
 //      for (m=1;m<=2*(n-1);m<<=1);
 //    int L=0,x=m; while (x>>=1) L++;
 //    for (int i=1;i<=m;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
 //    for (int i=0; i<m; i++) printf("%d ",rev[i]); cout<<endl;
     for (int i=; i<; i++) ntt[i].pre(M[i],G[i],m);
     for (int i=; i<; i++)
     {
         memcpy(c,a,sizeof(int)*(m+)); memcpy(d,b,sizeof(int)*(m+));
         ntt[i].FFT(c,m,); ntt[i].FFT(d,m,);
         for (int j=; j<m; j++) c[j]=(LL)c[j]*d[j]%ntt[i].P;
         ntt[i].FFT(c,m,);
         for (int j=; j<m; j++) ans[i][j]=c[j];
     }
     for (int i=; i<n; i++) printf("%d ",CRT(ans[][i],ans[][i],ans[][i]));
     return ;
 }