2
4
2016
0

【bzoj3160】万径人踪灭

3160: 万径人踪灭

Time Limit: 10 Sec  Memory Limit: 256 MB
Submit: 574  Solved: 332
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Description

Input

Output

 

 

Sample Input

 

Sample Output

 

HINT

 

 

Source

%%%PoPoQQQ,http://blog.csdn.net/PoPoQQQ/article/details/42193259

代码:

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define mod 1000000007
using namespace std;
double pi=acos(-1);
struct cp{double x,y;};
int n,m,ans,len;
int f[500010],mi[500010];
cp a[500010],b[500010],c[500010],cur[500010];
char s[500010],s1[500010];
cp operator *(cp x,cp y){return (cp){x.x*y.x-x.y*y.y,x.y*y.x+x.x*y.y};}
cp operator +(cp x,cp y){return (cp){x.x+y.x,x.y+y.y};}
cp operator -(cp x,cp y){return (cp){x.x-y.x,x.y-y.y};}
void fft(cp *a,int n,int fl)
{
    for (int i=n>>1,j=1;j<n;j++)
    {
        if (i<j) swap(a[i],a[j]);
        int k=n>>1;
        for (;k&i;i^=k,k>>=1);i^=k;
    }
    for (int m=2;m<=n;m<<=1)
    {
        cp w=(cp){cos(2*pi*fl/m),sin(2*pi*fl/m)};
        cur[0]=(cp){1,0};
        for (int i=1;i<=m;i++) cur[i]=cur[i-1]*w;
        for (int i=0;i<n;i+=m)
            for (int j=i;j<i+(m>>1);j++)
            {
                cp u=a[j],v=a[j+(m>>1)]*cur[j-i];
                a[j]=u+v;
                a[j+(m>>1)]=u-v;
            }
    }
}
int manacher()
{
    int ans=0;
    s1[0]='$';len=1;
    for (int i=0;i<n;i++){s1[len++]='#',s1[len++]=s[i];}
    s1[len++]='#';
    int mx=1,id=1;
    for (int i=1;i<len;i++)
    {
        f[i]=min(f[id*2-i],mx-i);
        while (s1[i+f[i]]==s1[i-f[i]]) f[i]++;
        if (f[i]+i>mx){mx=f[i]+i;id=i;}
        ans+=f[i]>>1;ans%=mod;
    }
    return ans;
}
int main()
{
    mi[0]=1;
    for (int i=1;i<=500000;i++) mi[i]=mi[i-1]*2%mod;
    scanf("%s",s);
    n=strlen(s);
    m=1;while (m<=n*2) m<<=1;
    for (int i=0;i<n;i++) if (s[i]=='a') a[i].x=1;
    fft(a,m,1);
    for (int i=0;i<m;i++) b[i]=a[i]*a[i];
    fft(b,m,-1);
    for (int i=0;i<m;i++) b[i].x=(int)(b[i].x/m+0.5);
    for (int i=0;i<m;i++) a[i].x=a[i].y=0;
    for (int i=0;i<n;i++) if (s[i]=='b') a[i].x=1;
    fft(a,m,1);
    for (int i=0;i<m;i++) c[i]=a[i]*a[i];
    fft(c,m,-1);
    for (int i=0;i<m;i++) c[i].x=(int)(c[i].x/m+0.5);
    for (int i=0;i<m;i++) f[i]=((int)b[i].x+(int)c[i].x+1)>>1;
    for (int i=0;i<m;i++) {ans+=(mi[f[i]]+mod-1)%mod;ans%=mod;}
    ans=ans+mod-manacher();ans%=mod;
    printf("%d\n",ans);
}
Category: FFT | Tags: | Read Count: 439

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