洛谷傳送門

解題分析

很容易想到： 方案數$=$迴文序列個數-迴文串個數。

迴文串我們用$manacher$搞出來就好了， 關鍵是迴文序列數。

聯絡到我們寫萬用字元匹配時的套路， 如果第$i$個字元和第$j$個字元相同，我們以$\frac{i+1}{2}$為中心情況的貢獻$cnt[i]$就會$+1$。

這個分數很討厭， 我們就把這個貢獻加在$i+j$就好， 最後答案就為$\sum_{i=1}^{n\times 2}(2^{cnt[i]}-1)$

$cnt[]$這玩意顯然是個卷積的形式， 可以通過自己和自己$FFT$得到， 完美$AC$。

注意計算的時候加上$EPS$， 誤差較大。

程式碼如下：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <algorithm>
#define R register
#define IN inline
#define
 
 W while
#define gc getchar()
#define MX 800500
#define db double
#define MOD 1000000007
#define EPS 0.1
const db PI = std::acos(-1.0);
struct Complex {db im, re;} a[MX], b[MX];
IN Complex operator * (const Complex &x, const Complex &y) {return {x.im * y.re + x.re * y.im, x.re * y.re - x.im * y.im};}
IN Complex operator
 
 + (const Complex &x, const Complex &y) {return {x.im + y.im, x.re + y.re};}
IN Complex operator - (const Complex &x, const Complex &y) {return {x.im - y.im, x.re - y.re};}
int rev[MX], ans[MX], pw[MX], cnt[MX];
int len, mid, pos, tot, lg, res;
char buf[MX], val[MX];
IN void FFT(Complex *dat, const int &typ)
{
	for (R int i = 0; i < tot; ++i) if(rev[i] > i) std::swap(dat[rev[i]], dat[i]);
	R int seg, bd, now, cur, step;
	Complex base, deal, buf1, buf2;
	for (seg = 1; seg < tot; seg <<= 1)
	{
		base = {std::sin(PI / seg) * typ, std::cos(PI / seg)}; step = seg << 1;
		for (now = 0; now < tot; now += step)
		{
			deal = {0, 1}, bd = now + seg;
			for (cur = now; cur < bd; ++cur, deal = deal * base)
			{
				buf1 = dat[cur], buf2 = dat[cur + seg] * deal;
				dat[cur] = buf1 + buf2, dat[cur + seg] = buf1 - buf2;
			}
		}
	}
}
IN int manacher()
{
	int ret = 0, bd = (len << 1) + 1;
	val[0] = '$'; val[1] = '#';
	for (R int i = 2; i <= bd; ++i)
	if(i & 1) val[i] = '#'; else val[i] = buf[(i >> 1) - 1];
	mid = 0, pos = 0;
	for (R int i = 1; i <= bd; ++i)
	{
		if(i < pos) ans[i] = std::min(ans[2 * mid - i], pos - i);
		else ans[i] = 1;
		W (val[i + ans[i]] == val[i - ans[i]]) ++ans[i];
		if(pos < ans[i] + i) mid = i, pos = ans[i] + i;
		ret = (ret + ans[i] / 2) % MOD;
	}
	return ret;
}
int main(void)
{
	scanf("%s", buf); len = std::strlen(buf);
	for (tot = pw[0] = 1; tot < len; tot <<= 1, ++lg); ++lg, tot <<= 1;
	for (R int i = 1; i < tot; ++i) pw[i] = pw[i - 1] * 2 % MOD;
	for (R int i = 1; i < tot; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lg - 1));
	for (R int i = 0; i < len; ++i) (buf[i] == 'a') ? (a[i].re = 1) : (b[i].re = 1);
	FFT(a, 1), FFT(b, 1);
	for (R int i = 0; i < tot; ++i) a[i] = a[i] * a[i], b[i] = b[i] * b[i];
	FFT(a, -1), FFT(b, -1);
	for (R int i = 0; i < tot; ++i) cnt[i] = (((int)((a[i].re + b[i].re) / tot + EPS) + 1) / 2);
	for (R int i = 0; i < tot; ++i) res = (res + pw[cnt[i]] - 1) % MOD;
	printf("%d", (res - manacher() + MOD) % MOD);
}