BZOJ3160【万径人踪灭】 【FFT】

。。恩 打了四五遍 不会也背出来了。。

BZOJ3160 【听说时限紧？转C++的优势么？】

上AC代码 fft

 /*Problem: 3160
     User: cyz666
     Language: C++
     Result: Accepted
     Time:1992 ms
     Memory:18492 kb
 ****************************************************************/

 #include <bits/stdc++.h>
 #define LL long long
 const LL mo=;
 const double pi=acos(-1.0);
 using namespace std;
 struct cp{
     double x,y;
     cp(double a=,double b=):x(a),y(b){}
     //cp(double a=0,double b=0){x=a,y=b;}
 }a[],b[],W[];
 int h[],rev[],f[],c[],n,x,m;
 char s; LL ans;
 cp operator +(cp a,cp b){
     cp c(a.x+b.x,a.y+b.y);
     return c;
 }
 cp operator *(cp a,cp b){
     cp c(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
     return c;
 }
 cp operator -(cp a,cp b){
     return (cp){a.x-b.x,a.y-b.y};
 }
 void fft(cp a[],int n,int d=){
     if (d<) reverse(a+,a+n);
     for (int i=;i<n;++i)
         if (i<rev[i]) swap(a[i],a[rev[i]]);
     for (int i=;i<n/;++i)
         W[i]=cp(cos(*pi/n*i),sin(*pi/n*i));
     for (int m=,k=;m<=n;k=m,m<<=)
         for (int i=;i<n;i+=m)
         for (int j=i;j<i+k;++j){
             cp u=a[j],v=a[j+k]*W[n/m*(j-i)];
             a[j]=u+v,a[j+k]=u-v;
         }
     if (d<) for (int i=;i<n;++i) a[i].x/=n;
 }
 int main(){
     while (scanf("%c",&s)&&isalpha(s))
         s=='a'?c[++++n]=:c[++++n]=;
     c[]=-; c[n+]=-;
     for (int i=;i<=n;++i){
         if (x+f[x]>i) f[i]=min(f[x+x-i],x+f[x]-i);
         while (c[i+f[i]]==c[i-f[i]]) ++f[i]; --f[i];
         if (i+f[i]>x+f[x]) x=i;
         ans+=f[i]/; if (ans>=mo) ans-=mo;
     }
     m=ceil((log(n)/log())); m=<<m;
     for (int i=;i<m;++i)rev[i]=(rev[i>>]>>)+(m>>)*(i&);
     x=; for (int i=;i<=n;++++i) c[++x]=c[i]; n=x;
     for (int i=;i<=n;++i) c[i]==?a[i].x=:b[i].x=;
     fft(a,m);  fft(b,m);
     for (int i=;i<m;++i) a[i]=a[i]*a[i],b[i]=b[i]*b[i];
     fft(a,m,-); fft(b,m,-);
     h[]=; for (int i=;i<=n;++i)h[i]=(h[i-]+h[i-])%mo;
     int x;  ans=mo-ans;
     for (int i=;i<n+n;++i){
         x=round(a[i].x+b[i].x);
         ans+=h[x+>>]-; if (ans>=mo) ans-=mo;
     }
     ans=(ans-n++mo)%mo;
     printf("%lld",ans);
     return ;
 }

KriTo

然后 顺手写了个ntt板子扔这（没用题目测过对不对）

mo的原根g的定义：g^1,g^2,...,g^(mo-1) 在%mo意义下各不相同。

 #include <bits/stdc++.h>
 #define LL long long
 using namespace std;
 int g1,g2,N,n,k,x,W[],a[],b[],rev[],mo1,mo2;
 LL po(LL x,LL y,LL mo){
     LL z=; if (y<) y=y%(mo-)+mo-;
     for (;y;y>>=,x=x*x%mo)
     if (y&) z=z*x%mo;
     return z;
 }
 int getg(LL mo){
     LL y=mo-,x=floor(sqrt(y)); int fl;
     for (int g=;;++g){
         fl=;
         for (int i=;i<=x;++i) if (!(y%i))
         if (po(g,i,mo)==||po(g,y/i,mo)==) {fl=;break;}
         if (fl) return g;
     }
 }
 void ntt(int a[],int n,int d,int mo,int G){    //n整除(mo-1)
     if (d<) reverse(a+,a+n);
     for (int i=;i<n;++i)
         if (i<rev[i]) swap(a[i],a[rev[i]]);
     W[]=; LL X=po(G,(mo-)/n,mo);
     for (int i=;i<n/;++i) W[i]=X*W[i-]%mo;
     for (int m=,k=;m<=n;k=m,m<<=)
         for (int i=;i<n;i+=m)
         for (int j=i;j<i+k;++j){
             int u=a[j],v=1ll*a[j+k]*W[n/m*(j-i)]%mo;
             a[j]=u+v>=mo?u+v-mo:u+v;
             a[j+k]=u-v<?u-v+mo:u-v;
         }
     if (d<){
         X=po(n,mo-,mo);
         for (int i=;i<n;++i) a[i]=X*a[i]%mo;
     }
 }
 int main(){
     mo1=; mo2=;
     g1=getg(mo1); g2=getg(mo2);
     scanf("%d%d",&n,&k); W[]=;
     for (int i=;i<=n;++i)
         scanf("%d",&x),a[x]=,b[x]=;
     N=;
     for (int i=;i<N;++i) rev[i]=rev[i>>]>>|(i&?N>>:);
     ntt(a,N,,mo1,g1); ntt(b,N,,mo2,g2);
     for (int i=;i<N;++i)
         a[i]=po(a[i],k,mo1),b[i]=po(b[i],k,mo2);
     ntt(a,N,-,mo1,g1); ntt(b,N,-,mo2,g2);
     for (int i=;i<N;++i){
         if (a[i]<) a[i]+=mo1;
         if (b[i]<) b[i]+=mo2;
     }
     for (int i=;i<N;++i) if (a[i]||b[i]) printf("%d ",i); puts("");
     return ;
 }

AsuNa

若多项式相乘的最高次为n， 则FFT,NTT的数组大小要开到>n的最小的2的幂。 即 1<<(int)ceil(log(n+1)/log(2))

给几个模数：

1004535809=479*2^21+1,   原根=3

998244353=7*17*2^23+1 ，原根=3

469762049=7*2^26+1      ，原根=3