BZOJ3160 万径人踪灭（FFT+manacher）

　　容易想到先统计回文串数量，这样就去掉了不连续的限制，变为统计回文序列数量。

　　显然以某个位置为对称轴的回文序列数量就是2^{其两边（包括自身）对称相等的位置数量}-1。对称有啥性质？位置和相等。这不就是卷积嘛。那么就做完了。

　　又写挂manacher，没救。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 270000
#define P 1000000007
int n,m,s[N],t,r[N],f[N],p[N],ans=;
const double PI=3.14159265358979324;
struct complex
{
    double x,y;
    complex operator +(const complex&a) const
    {
        return (complex){x+a.x,y+a.y};
    }
    complex operator -(const complex&a) const
    {
        return (complex){x-a.x,y-a.y};
    }
    complex operator *(const complex&a) const
    {
        return (complex){x*a.x-y*a.y,x*a.y+y*a.x};
    }
}a[N],b[N];
void DFT(int n,complex *a,int p)
{
    for (int i=;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
    for (int i=;i<=n;i<<=)
    {
        complex wn=(complex){cos(*PI/i),p*sin(*PI/i)};
        for (int j=;j<n;j+=i)
        {
            complex w=(complex){,};
            for (int k=j;k<j+(i>>);k++,w=w*wn)
            {
                complex x=a[k],y=w*a[k+(i>>)];
                a[k]=x+y,a[k+(i>>)]=x-y;
            }
        }
    }
}
void mul(int n,complex *a,complex *b)
{
    for (int i=;i<n;i++) a[i].y=a[i].x-b[i].x,a[i].x=a[i].x+b[i].x;
    DFT(n,a,);
    for (int i=;i<n;i++) a[i]=a[i]*a[i];
    DFT(n,a,-);
    for (int i=;i<n;i++) a[i].x=a[i].x/n/;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj3160.in","r",stdin);
    freopen("bzoj3160.out","w",stdout);
    const char LL[]="%I64d";
#else
    const char LL[]="%lld";
#endif
    char c=getchar();
    while (c=='a'||c=='b') s[n++]=c-,c=getchar();
    m=n*-;
    t=;while (t<m) t<<=;
    for (int i=;i<t;i++) r[i]=(r[i>>]>>)|(i&)*(t>>);
    for (int i=;i<n;i++) a[i].x=b[i].x=s[i]-;
    mul(t,a,b);
    for (int i=;i<m;i++) f[i]=(int){a[i].x+0.5};
    for (int i=;i<t;i++) a[i].x=a[i].y=b[i].x=b[i].y=;
    for (int i=;i<n;i++) a[i].x=b[i].x=-s[i];
    mul(t,a,b);
    for (int i=;i<m;i++) f[i]+=(int){a[i].x+0.5};
    for (int i=;i<m;i++) f[i]=f[i]+>>;
    p[]=;
    for (int i=;i<m;i++) p[i]=(p[i-]<<)%P;
    for (int i=;i<m;i++) ans=(ans+p[f[i]]-)%P;
    for (int i=m-;~i;i--) s[i]=(i&?:s[i>>]);
    for (int i=m;i;i--) s[i]=s[i-];
    s[]=s[++m]=;
    memset(r,,sizeof(r));
    r[]=;int x=;
    for (int i=;i<=m;i++)
    {
        if (x+r[x]>=i) r[i]=min(r[x-(i-x)],x+r[x]-i);
        while (i-r[i]->=&&i+r[i]+<=m&&s[i+r[i]+]==s[i-r[i]-]) r[i]++;
        if (i+r[i]>=x+r[x]) x=i;
    }
    for (int i=;i<=m;i++) ans=(ans-(r[i]+>>)+P)%P;
    cout<<ans;
    return ;
}