bzoj3160: 万径人踪灭

数组开小gg

fft乱搞就可以了。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const LL mod=1e9+7; 
const double pi=acos(-1.0);struct complex
{double r,i;complex(){}complex(double R,double I){r=R,i=I;}friend complex operator +(complex x,complex y){return complex(x.r+y.r,x.i+y.i);}friend complex operator -(complex x,complex y){return complex(x.r-y.r,x.i-y.i);}friend complex operator *(complex x,complex y){return complex(x.r*y.r-x.i*y.i,x.r*y.i+x.i*y.r);}
}A[410000]; int Re[410000];
void fft(complex *a,int n,int op)
{for(int i=0;i<n;i++)if(i<Re[i])swap(a[i],a[Re[i]]);for(int i=1;i<n;i<<=1){complex wn(cos(pi/i),sin(op*pi/i));for(int j=0;j<n;j+=(i<<1)){complex w(1,0);for(int k=0;k<i;k++,w=w*wn){complex t1=a[j+k],t2=a[j+k+i]*w;a[j+k]=t1+t2;a[j+k+i]=t1-t2;}}}
}int a[410000],d[410000];
void solve(int len)
{len++;memset(A,0,sizeof(A));for(int i=0;i<len;i++)A[i].r=double(a[i]);int n,m=len*2,L=0;for(n=1;n<=m;n*=2)L++;for(int i=1;i<=n;i++)Re[i]=(Re[i>>1]>>1)|((i&1)<<(L-1));fft(A,n,1);for(int i=0;i<=n;i++)A[i]=A[i]*A[i];fft(A,n,-1);for(int i=0;i<=n;i++)d[i]+=(int(A[i].r/double(n)+0.5)+1)/2;
}char ss[410000];
int sa[410000],p[410000];
LL manacher(int n)
{int len=0; sa[++len]=0;for(int i=1;i<=n;i++)sa[++len]=ss[i]-'a'+1, sa[++len]=0;int k=1;p[1]=1;for(int i=2;i<=len;i++){int L=k-p[k]+1,R=k+p[k]-1;int j=k-(i-k);if(R>=i)p[i]=min(p[j],j-L+1);else p[i]=0;while(i-p[i]>0&&i+p[i]<=len&&sa[i+p[i]]==sa[i-p[i]])p[i]++;if(i+p[i]-1>k+p[k]-1)k=i;}LL ret=0;for(int i=1;i<=len;i++)if(i%2==1)ret=(ret+(p[i]-1)/2)%mod;else ret=(ret+p[i]/2)%mod;return ret;
}LL quick_pow(LL A,LL p) 
{LL ret=1;while(p!=0) {if(p%2==1)ret=ret*A%mod;A=A*A%mod;p/=2;}return ret; 
}
int main()
{freopen("a.in","r",stdin);freopen("a.out","w",stdout);scanf("%s",ss+1); int n=strlen(ss+1);LL ddd=manacher(n); a[0]=0;for(int i=1;i<=n;i++)a[i]=ss[i]-'a'; solve(n);for(int i=1;i<=n;i++)a[i]^=1;         solve(n);LL sum=0;for(int i=2;i<=n*2;i++){sum=(sum+quick_pow(2,d[i])-1+mod)%mod;}printf("%lld\n",((sum-ddd)%mod+mod)%mod);return 0;
}