题面
Sol
求不连续回文子序列的个数
(ans=)回文子序列个数-连续回文子序列个数
即回文子序列个数-回文子串个数
后面直接(Manacher)就好了
考虑前面的
枚举对称轴，设(f[i])表示对称轴(i)两边相同字符的对数
那么最终答案就是(sum 2^{f[i]}-1)
考虑求(f[i])
只有当原串中的两个字符相同才会有贡献
也就是(s[i-x]=s[i+x])
单独考虑(a)和(b)的贡献
(f[i]=sum [s[i-x]==s[i+x]])
设当前考虑(a)的贡献
把是(a)的设为(1)不是的为(0)，做个卷积就可以求出(f)
那么就可以(FFT)辣
注意两个字符之间也算对称轴
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int Zsy(1e9 + 7);
const int _(5e5 + 5);
const double PI(acos(-1));
int n, p[_];
ll f[_];
int N, M, l, r[_];
char s[_], a[_];
struct Complex{
double real, image;
IL Complex(){
real = image = 0;
}
IL Complex(RG double a, RG double b){
real = a, image = b;
}
IL Complex operator +(RG Complex B){
return Complex(real + B.real, image + B.image);
}
IL Complex operator -(RG Complex B){
return Complex(real - B.real, image - B.image);
}
IL Complex operator *(RG Complex B){
return Complex(real * B.real - image * B.image, real * B.image + image * B.real);
}
} A[_], B[_];
IL void FFT(RG Complex *P, RG int opt){
for(RG int i = 0; i < N; ++i) if(i < r[i]) swap(P[i], P[r[i]]);
for(RG int i = 1; i < N; i <<= 1){
RG Complex W(cos(PI / i), opt * sin(PI / i));
for(RG int j = 0, p = i << 1; j < N; j += p){
RG Complex w(1, 0);
for(RG int k = 0; k < i; ++k, w = w * W){
RG Complex X = P[k + j], Y = w * P[k + j + i];
P[k + j] = X + Y, P[k + j + i] = X - Y;
}
}
}
}
IL void Mul(){
FFT(A, 1);
for(RG int i = 0; i < N; ++i) B[i] = A[i] * A[i];
FFT(B, -1);
for(RG int i = 0; i < N; ++i) B[i].real = B[i].real / N + 0.5;
for(RG int i = 1; i <= M; ++i) f[i] += ((ll)(B[i].real) + 1) >> 1;
}
IL ll Manacher(){
RG ll ans = 0; RG int mx = 0, len = 1; a[1] = '#';
for(RG int i = 1; i <= n; ++i) a[++len] = s[i], a[++len] = '#';
for(RG int i = 1, id = 0, mx = 0; i <= len; ++i){
if(i < mx) p[i] = min(mx - i, p[(id << 1) - i]);
while(i - p[i] && i + p[i] <= len && a[i - p[i]] == a[i + p[i]]) ++p[i];
if(p[i] + i > mx) mx = p[i] + i, id = i;
(ans += p[i] >> 1) %= Zsy;
}
return ans;
}
IL ll Pow(RG ll x, RG ll y){
RG ll ret = 1;
for(; y; y >>= 1, x = x * x % Zsy)
if(y & 1) ret = ret * x % Zsy;
return ret;
}
int main(RG int argc, RG char* argv[]){
scanf(" %s", s + 1); n = strlen(s + 1);
for(M = n + n, N = 1; N <= M; N <<= 1) ++l;
for(RG int i = 0; i < N; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
RG ll ans = -Manacher();
for(RG int i = 0; i < N; ++i) A[i] = Complex(s[i] == 'a', 0);
Mul();
for(RG int i = 0; i < N; ++i) A[i] = Complex(s[i] == 'b', 0);
Mul();
for(RG int i = 1; i <= M; ++i) (ans += Pow(2, f[i]) - 1) % Zsy;
printf("%lld
", (ans + Zsy) % Zsy);
return 0;
}