# 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 x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
#define N 270000
#define P 1000000007
int n,m,s[N],t,r[N],f[N],p[N],ans=0;
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=0;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
for (int i=2;i<=n;i<<=1)
{
complex wn=(complex){cos(2*PI/i),p*sin(2*PI/i)};
for (int j=0;j<n;j+=i)
{
complex w=(complex){1,0};
for (int k=j;k<j+(i>>1);k++,w=w*wn)
{
complex x=a[k],y=w*a[k+(i>>1)];
a[k]=x+y,a[k+(i>>1)]=x-y;
}
}
}
}
void mul(int n,complex *a,complex *b)
{
for (int i=0;i<n;i++) a[i].y=a[i].x-b[i].x,a[i].x=a[i].x+b[i].x;
DFT(n,a,1);
for (int i=0;i<n;i++) a[i]=a[i]*a[i];
DFT(n,a,-1);
for (int i=0;i<n;i++) a[i].x=a[i].x/n/4;
}
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-96,c=getchar();
m=n*2-1;
t=1;while (t<m) t<<=1;
for (int i=0;i<t;i++) r[i]=(r[i>>1]>>1)|(i&1)*(t>>1);
for (int i=0;i<n;i++) a[i].x=b[i].x=s[i]-1;
mul(t,a,b);
for (int i=0;i<m;i++) f[i]=(int){a[i].x+0.5};
for (int i=0;i<t;i++) a[i].x=a[i].y=b[i].x=b[i].y=0;
for (int i=0;i<n;i++) a[i].x=b[i].x=2-s[i];
mul(t,a,b);
for (int i=0;i<m;i++) f[i]+=(int){a[i].x+0.5};
for (int i=0;i<m;i++) f[i]=f[i]+1>>1;
p[0]=1;
for (int i=1;i<m;i++) p[i]=(p[i-1]<<1)%P;
for (int i=0;i<m;i++) ans=(ans+p[f[i]]-1)%P;
for (int i=m-1;~i;i--) s[i]=(i&1?0:s[i>>1]);
for (int i=m;i;i--) s[i]=s[i-1];
s[0]=s[++m]=0;
memset(r,0,sizeof(r));
r[0]=0;int x=0;
for (int i=1;i<=m;i++)
{
if (x+r[x]>=i) r[i]=min(r[x-(i-x)],x+r[x]-i);
while (i-r[i]-1>=0&&i+r[i]+1<=m&&s[i+r[i]+1]==s[i-r[i]-1]) r[i]++;
if (i+r[i]>=x+r[x]) x=i;
}
for (int i=0;i<=m;i++) ans=(ans-(r[i]+1>>1)+P)%P;
cout<<ans;
return 0;
}

posted @ 2018-08-10 15:19  Gloid  阅读(122)  评论(0编辑  收藏  举报