BZOJ 3160 FFT+马拉车

题意显然

ans=回文子序列数目 - 回文子串数目 

回文子串直接用马拉车跑出来

回文子序列一开始总是不知道怎么求 （太蠢了）

后面看了题解

构造一个神奇的卷积

（这个是我盗的图）地址 

后面还有一些细节需要处理出 f[x] （f[x] 表示 x左右相等的个数）

通常我们需要的情况是 两个函数相乘 这里是s[x-i] == s[x+i] 分类讨论就行了  变成1*1=1的形式

所以要a=1 b=0 和 a=0 b=1都算一次

这里长度扩展了一倍 表示 当 i 是奇数时表示对称轴是元素 ，偶数表示对称轴是两个元素的间隔

根据二项式定理 求出每一个f[x] 的贡献 expmod ( 2, ( cnt[i] + 1 ) >> 1 ) - 1 

还有最后一个细节 相减的时候要记得加上mod 再取模

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define  pi acos(-1.0)
#define  eps 1e-9
#define  fi first
#define  se second
#define  rtl   rt<<1
#define  rtr   rt<<1|1
#define  bug         printf("******
")
#define  mem(a,b)    memset(a,b,sizeof(a))
#define  name2str(x) #x
#define  fuck(x)     cout<<#x" = "<<x<<endl
#define  f(a)        a*a
#define  sf(n)       scanf("%d", &n)
#define  sff(a,b)    scanf("%d %d", &a, &b)
#define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  pf          printf
#define  FRE(i,a,b)  for(i = a; i <= b; i++)
#define  FREE(i,a,b) for(i = a; i >= b; i--)
#define  FRL(i,a,b)  for(i = a; i < b; i++)+
#define  FRLL(i,a,b) for(i = a; i > b; i--)
#define  FIN         freopen("data.txt","r",stdin)
#define  gcd(a,b)    __gcd(a,b)
#define  lowbit(x)   x&-x
#define rep(i,a,b) for(int i=a;i<b;++i)
#define per(i,a,b) for(int i=a-1;i>=b;--i)
using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
const int maxn = 3e5 + 7;
const int maxm = maxn * 4;
const int mod = 1e9 + 7;
int n, m, a[maxn], b[maxn];
int len, res[maxm], mx; //开大4倍
struct cpx {
    long double r, i;
    cpx ( long double r = 0, long double i = 0 ) : r ( r ), i ( i ) {};
    cpx operator+ ( const cpx &b ) {
        return cpx ( r + b.r, i + b.i );
    }
    cpx operator- ( const cpx &b ) {
        return cpx ( r - b.r, i - b.i );
    }
    cpx operator* ( const cpx &b ) {
        return cpx ( r * b.r - i * b.i, i * b.r + r * b.i );
    }
} va[maxm], vb[maxm];
void rader ( cpx F[], int len ) { //len = 2^M,reverse F[i] with  F[j] j为i二进制反转
    int j = len >> 1;
    for ( int i = 1; i < len - 1; ++i ) {
        if ( i < j ) swap ( F[i], F[j] ); // reverse
        int k = len >> 1;
        while ( j >= k ) j -= k, k >>= 1;
        if ( j < k ) j += k;
    }
}
void FFT ( cpx F[], int len, int t ) {
    rader ( F, len );
    for ( int h = 2; h <= len; h <<= 1 ) {
        cpx wn ( cos ( -t * 2 * pi / h ), sin ( -t * 2 * pi / h ) );
        for ( int j = 0; j < len; j += h ) {
            cpx E ( 1, 0 ); //旋转因子
            for ( int k = j; k < j + h / 2; ++k ) {
                cpx u = F[k];
                cpx v = E * F[k + h / 2];
                F[k] = u + v;
                F[k + h / 2] = u - v;
                E = E * wn;
            }
        }
    }
    if ( t == -1 ) //IDFT
        for ( int i = 0; i < len; ++i ) F[i].r /= len;
}
void Conv ( cpx a[], cpx b[], int len ) { //求卷积
    FFT ( a, len, 1 );
    FFT ( b, len, 1 );
    for ( int i = 0; i < len; ++i ) a[i] = a[i] * b[i];
    FFT ( a, len, -1 );
}
void gao () {
    len = 1;
    mx = n + m;
    while ( len <= mx ) len <<= 1; //mx为卷积后最大下标
    for ( int i = 0; i < len; i++ ) va[i].r = va[i].i = vb[i].r = vb[i].i = 0;
    for ( int i = 0; i < n; i++ ) va[i].r = a[i]; //根据题目要求改写
    for ( int i = 0; i < m; i++ ) vb[i].r = b[i]; //根据题目要求改写
    Conv ( va, vb, len );
    for ( int i = 0; i < len; ++i ) res[i] += ( LL ) floor ( va[i].r + 0.5 );
}
char Ma[maxn * 2];
int Mp[maxn * 2];
int Manacher ( char s[], int len ) {
    int l = 0, ret = 0;
    Ma[l++] = '$';
    Ma[l++] = '#';
    for ( int i = 0; i < len; i++ ) {
        Ma[l++] = s[i];
        Ma[l++] = '#';
    }
    Ma[l] = 0;
    int mx = 0, id = 0;
    for ( int i = 0; i < l; i++ ) {
        Mp[i] = mx > i ? min ( Mp[2 * id - i], mx - i ) : 1;
        while ( Ma[i + Mp[i]] == Ma[i - Mp[i]] ) Mp[i]++;
        if ( i + Mp[i] > mx ) {
            mx = i + Mp[i];
            id = i;
        }
        ret += Mp[i] >> 1, ret %= mod;
    }
    return ret % mod;
}
LL expmod ( LL a, LL b ) {
    LL res = 1;
    while ( b ) {
        if ( b & 1 )  res = res * a % mod;
        a = a * a % mod;
        b = b >> 1;
    }
    return res % mod;
}
char s[maxn];
int cnt[maxn];
int main() {
   // FIN;
    scanf ( "%s", s + 1 );
    n = m = strlen ( s + 1 );
    LL ans = 0, temp = Manacher ( s + 1, n );
    n++, m++;
    for ( int i = 1 ; i < n  ; i++ ) if ( s[i] == 'a' ) a[i] = 1, b[i] = 1;
    gao();
    for ( int i = 1 ; i <= 2 * ( n - 1 ) ; i++ ) cnt[i] += res[i];
    mem ( a, 0 ), mem ( b, 0 ), mem ( res, 0 );
    for ( int i = 1 ; i < n  ; i++ ) if ( s[i] == 'b' ) a[i] = 1, b[i] = 1;
    gao();
    for ( int i = 1 ; i <= 2 * ( n - 1 ) ; i++ ) cnt[i] += res[i];
    for ( int i = 1 ; i <= 2 * ( n - 1 ) ; i++ )
        ans = ( ans + expmod ( 2, ( cnt[i] + 1 ) >> 1 ) - 1 ) % mod;
    printf ( "%lld
", ( ans - temp + mod ) % mod );
    return 0;
}