# 3160: 万径人踪灭

f[i]，i为奇数表示以缝隙对称，偶数表示以元素i>>1对称，对答案的贡献就是$2^{f[i]}-1$

$f[i] = \sum_{j=1}^{i-1} [s_j = s_{i-j}]$

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define fir first
#define sec second
typedef long long ll;
const int N=(1<<19)+5, mo=1e9+7;
const double PI = acos(-1);
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
return x*f;
}

struct meow{
double x, y;
meow(double a=0, double b=0):x(a), y(b){}
};
meow operator +(meow a, meow b) {return meow(a.x+b.x, a.y+b.y);}
meow operator -(meow a, meow b) {return meow(a.x-b.x, a.y-b.y);}
meow operator *(meow a, meow b) {return meow(a.x*b.x-a.y*b.y, a.x*b.y+a.y*b.x);}
meow conj(meow a) {return meow(a.x, -a.y);}
typedef meow cd;

namespace fft {
int n, rev[N];
void ini(int lim) {
n=1; int k=0; while(n<lim) n<<=1, k++;
for(int i=0; i<n; i++) rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));
}
void dft(cd *a, int flag) {
for(int i=0; i<n; i++) if(i<rev[i]) swap(a[i], a[rev[i]]);
for(int l=2; l<=n; l<<=1) {
int m = l>>1;
cd wn = cd(cos(2*PI/l), flag * sin(2*PI/l));
for(cd *p=a; p != a+n; p+=l) {
cd w(1, 0);
for(int k=0; k<m; k++) {
cd t = w * p[k+m];
p[k+m] = p[k] - t;
p[k] = p[k] + t;
w = w*wn;
}
}
}
if(flag == -1) for(int i=0; i<n; i++) a[i].x /= n;
}
void mul(cd *a) {
dft(a, 1);
for(int i=0; i<n; i++) a[i] = a[i] * a[i];
dft(a, -1);
}
}

cd a[N];
int n, f[N]; ll ans;
char s[N];
ll Pow(ll a, int b) {
ll ans=1;
for(; b; b>>=1, a=a*a%mo)
if(b&1) ans=ans*a%mo;
return ans;
}
inline void mod(ll &x) {if(x<0) x+=mo; else if(x>=mo) x-=mo;}

namespace ma {
int r[N]; char a[N];
ll manacher(char *s, int n) {
int p=0, a; ll ans=0;
for(int i=1; i<=n; i++) {
r[i] = i<p ? min(p-i+1, r[2*a-i]) : 1;
while(s[ i-r[i] ] == s[ i+r[i] ]) r[i]++;
if(i+r[i]-1 > p) p = i+r[i]-1, a=i;
mod(ans += r[i]>>1);
}
return ans;
}
ll cal(char *s) {
for(int i=1; i<=n; i++) a[(i<<1)-1] = '#', a[i<<1] = s[i];
a[(n<<1)+1] = '#';
a[0] = '@'; a[(n<<1)+2] = '\$';
return manacher(a, (n<<1)+1);
}
}

int main() {
freopen("in","r",stdin);
scanf("%s", s+1); n = strlen(s+1);
fft::ini(n+n+1);
for(int i=1; i<=n; i++) a[i].x = s[i]=='a';
fft::mul(a);
for(int i=1; i<=n+n; i++) f[i] = (int)floor(a[i].x + 0.5);

for(int i=0; i<fft::n; i++) a[i] = cd(0, 0);
for(int i=1; i<=n; i++) a[i].x = s[i]=='b';
fft::mul(a);
for(int i=1; i<=n+n; i++) f[i] += (int)floor(a[i].x + 0.5);
for(int i=1; i<=n+n; i++) f[i] = (f[i]+1)>>1;
//for(int i=1; i<=n+n; i++) printf("f %d  %d\n", i, f[i]);

for(int i=1; i<=n+n; i++) mod(ans += Pow(2, f[i]) - 1);

//printf("ans1 %lld\n", ans);
mod(ans -= ma::cal(s));
printf("%lld", ans);
}


posted @ 2017-04-12 12:13  Candy?  阅读(301)  评论(0编辑  收藏  举报