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=1000000007;
const double pi=acos(-1.0);
using namespace std;
struct cp{
    double x,y;
    cp(double a=0,double b=0):x(a),y(b){}
    //cp(double a=0,double b=0){x=a,y=b;}
}a[400005],b[400005],W[200005];
int h[400005],rev[400005],f[200005],c[200005],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=1){
    if (d<0) reverse(a+1,a+n);
    for (int i=0;i<n;++i)
        if (i<rev[i]) swap(a[i],a[rev[i]]);
    for (int i=0;i<n/2;++i)
        W[i]=cp(cos(2*pi/n*i),sin(2*pi/n*i));
    for (int m=2,k=1;m<=n;k=m,m<<=1)
        for (int i=0;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<0) for (int i=0;i<n;++i) a[i].x/=n;
}
int main(){
    while (scanf("%c",&s)&&isalpha(s))
        s=='a'?c[++++n]=1:c[++++n]=2;
    c[0]=-1; c[n+2]=-2;
    for (int i=1;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]/2; if (ans>=mo) ans-=mo;
    }
    m=ceil((log(n)/log(2))); m=1<<m;
    for (int i=1;i<m;++i)rev[i]=(rev[i>>1]>>1)+(m>>1)*(i&1);
    x=0; for (int i=2;i<=n;++++i) c[++x]=c[i]; n=x;
    for (int i=1;i<=n;++i) c[i]==1?a[i].x=1:b[i].x=1;
    fft(a,m);  fft(b,m);
    for (int i=0;i<m;++i) a[i]=a[i]*a[i],b[i]=b[i]*b[i];
    fft(a,m,-1); fft(b,m,-1);
    h[0]=1; for (int i=1;i<=n;++i)h[i]=(h[i-1]+h[i-1])%mo;
    int x;  ans=mo-ans; 
    for (int i=1;i<n+n;++i){
        x=round(a[i].x+b[i].x); 
        ans+=h[x+1>>1]-1; if (ans>=mo) ans-=mo;
    }
    ans=(ans-n+1+mo)%mo;
    printf("%lld",ans);
    return 0;
}

  

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[1050000],a[1050000],b[1050000],rev[1050000],mo1,mo2;
LL po(LL x,LL y,LL mo){
    LL z=1; if (y<0) y=y%(mo-1)+mo-1;
    for (;y;y>>=1,x=x*x%mo)
    if (y&1) z=z*x%mo;
    return z;
}
int getg(LL mo){
    LL y=mo-1,x=floor(sqrt(y)); int fl; 
    for (int g=2;;++g){
        fl=1;
        for (int i=2;i<=x;++i) if (!(y%i))
        if (po(g,i,mo)==1||po(g,y/i,mo)==1) {fl=0;break;}
        if (fl) return g;
    }
}
void ntt(int a[],int n,int d,int mo,int G){    //n整除(mo-1)
    if (d<0) reverse(a+1,a+n);
    for (int i=0;i<n;++i)
        if (i<rev[i]) swap(a[i],a[rev[i]]);
    W[0]=1; LL X=po(G,(mo-1)/n,mo);
    for (int i=1;i<n/2;++i) W[i]=X*W[i-1]%mo;
    for (int m=2,k=1;m<=n;k=m,m<<=1)
        for (int i=0;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<0?u-v+mo:u-v;
        }
    if (d<0){
        X=po(n,mo-2,mo);
        for (int i=0;i<n;++i) a[i]=X*a[i]%mo;
    }
}
int main(){
    mo1=998244353; mo2=1004535809;
    g1=getg(mo1); g2=getg(mo2);
    scanf("%d%d",&n,&k); W[0]=1;
    for (int i=1;i<=n;++i)
        scanf("%d",&x),a[x]=1,b[x]=1;
    N=1048576;
    for (int i=0;i<N;++i) rev[i]=rev[i>>1]>>1|(i&1?N>>1:0);
    ntt(a,N,1,mo1,g1); ntt(b,N,1,mo2,g2);
    for (int i=0;i<N;++i)
        a[i]=po(a[i],k,mo1),b[i]=po(b[i],k,mo2);
    ntt(a,N,-1,mo1,g1); ntt(b,N,-1,mo2,g2);
    for (int i=0;i<N;++i){
        if (a[i]<0) a[i]+=mo1;
        if (b[i]<0) b[i]+=mo2;
    }
    for (int i=0;i<N;++i) if (a[i]||b[i]) printf("%d ",i); puts("");
    return 0;
}

 

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