3160: 万径人踪灭

题意：求一个序列有多少不连续的回文子序列

一开始zz了直接用\(2^{r_i}-1\)

总-回文子串

后者用manacher处理

前者，考虑回文有两种对称形式(以元素/缝隙作为对称轴)

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\(2^{f[i]}-1\)
\[ f[i] = \sum_{j=1}^{i-1} [s_j = s_{i-j}] \]
就是裸卷积

因为只有a，b两个字符，可以先后令a或b=1分别求

#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);
inline int read(){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);
}