啥都不会啊!怎么办啊!

Fitz

慢慢来生活总会好起来的!!!

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

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <queue>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <set>
  7 #include <iostream>
  8 #include <map>
  9 #include <stack>
 10 #include <string>
 11 #include <vector>
 12 #define  pi acos(-1.0)
 13 #define  eps 1e-9
 14 #define  fi first
 15 #define  se second
 16 #define  rtl   rt<<1
 17 #define  rtr   rt<<1|1
 18 #define  bug         printf("******\n")
 19 #define  mem(a,b)    memset(a,b,sizeof(a))
 20 #define  name2str(x) #x
 21 #define  fuck(x)     cout<<#x" = "<<x<<endl
 22 #define  f(a)        a*a
 23 #define  sf(n)       scanf("%d", &n)
 24 #define  sff(a,b)    scanf("%d %d", &a, &b)
 25 #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
 26 #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
 27 #define  pf          printf
 28 #define  FRE(i,a,b)  for(i = a; i <= b; i++)
 29 #define  FREE(i,a,b) for(i = a; i >= b; i--)
 30 #define  FRL(i,a,b)  for(i = a; i < b; i++)+
 31 #define  FRLL(i,a,b) for(i = a; i > b; i--)
 32 #define  FIN         freopen("data.txt","r",stdin)
 33 #define  gcd(a,b)    __gcd(a,b)
 34 #define  lowbit(x)   x&-x
 35 #define rep(i,a,b) for(int i=a;i<b;++i)
 36 #define per(i,a,b) for(int i=a-1;i>=b;--i)
 37 using namespace std;
 38 typedef long long  LL;
 39 typedef unsigned long long ULL;
 40 const int maxn = 3e5 + 7;
 41 const int maxm = maxn * 4;
 42 const int mod = 1e9 + 7;
 43 int n, m, a[maxn], b[maxn];
 44 int len, res[maxm], mx; //开大4倍
 45 struct cpx {
 46     long double r, i;
 47     cpx ( long double r = 0, long double i = 0 ) : r ( r ), i ( i ) {};
 48     cpx operator+ ( const cpx &b ) {
 49         return cpx ( r + b.r, i + b.i );
 50     }
 51     cpx operator- ( const cpx &b ) {
 52         return cpx ( r - b.r, i - b.i );
 53     }
 54     cpx operator* ( const cpx &b ) {
 55         return cpx ( r * b.r - i * b.i, i * b.r + r * b.i );
 56     }
 57 } va[maxm], vb[maxm];
 58 void rader ( cpx F[], int len ) { //len = 2^M,reverse F[i] with  F[j] j为i二进制反转
 59     int j = len >> 1;
 60     for ( int i = 1; i < len - 1; ++i ) {
 61         if ( i < j ) swap ( F[i], F[j] ); // reverse
 62         int k = len >> 1;
 63         while ( j >= k ) j -= k, k >>= 1;
 64         if ( j < k ) j += k;
 65     }
 66 }
 67 void FFT ( cpx F[], int len, int t ) {
 68     rader ( F, len );
 69     for ( int h = 2; h <= len; h <<= 1 ) {
 70         cpx wn ( cos ( -t * 2 * pi / h ), sin ( -t * 2 * pi / h ) );
 71         for ( int j = 0; j < len; j += h ) {
 72             cpx E ( 1, 0 ); //旋转因子
 73             for ( int k = j; k < j + h / 2; ++k ) {
 74                 cpx u = F[k];
 75                 cpx v = E * F[k + h / 2];
 76                 F[k] = u + v;
 77                 F[k + h / 2] = u - v;
 78                 E = E * wn;
 79             }
 80         }
 81     }
 82     if ( t == -1 ) //IDFT
 83         for ( int i = 0; i < len; ++i ) F[i].r /= len;
 84 }
 85 void Conv ( cpx a[], cpx b[], int len ) { //求卷积
 86     FFT ( a, len, 1 );
 87     FFT ( b, len, 1 );
 88     for ( int i = 0; i < len; ++i ) a[i] = a[i] * b[i];
 89     FFT ( a, len, -1 );
 90 }
 91 void gao () {
 92     len = 1;
 93     mx = n + m;
 94     while ( len <= mx ) len <<= 1; //mx为卷积后最大下标
 95     for ( int i = 0; i < len; i++ ) va[i].r = va[i].i = vb[i].r = vb[i].i = 0;
 96     for ( int i = 0; i < n; i++ ) va[i].r = a[i]; //根据题目要求改写
 97     for ( int i = 0; i < m; i++ ) vb[i].r = b[i]; //根据题目要求改写
 98     Conv ( va, vb, len );
 99     for ( int i = 0; i < len; ++i ) res[i] += ( LL ) floor ( va[i].r + 0.5 );
100 }
101 char Ma[maxn * 2];
102 int Mp[maxn * 2];
103 int Manacher ( char s[], int len ) {
104     int l = 0, ret = 0;
105     Ma[l++] = '$';
106     Ma[l++] = '#';
107     for ( int i = 0; i < len; i++ ) {
108         Ma[l++] = s[i];
109         Ma[l++] = '#';
110     }
111     Ma[l] = 0;
112     int mx = 0, id = 0;
113     for ( int i = 0; i < l; i++ ) {
114         Mp[i] = mx > i ? min ( Mp[2 * id - i], mx - i ) : 1;
115         while ( Ma[i + Mp[i]] == Ma[i - Mp[i]] ) Mp[i]++;
116         if ( i + Mp[i] > mx ) {
117             mx = i + Mp[i];
118             id = i;
119         }
120         ret += Mp[i] >> 1, ret %= mod;
121     }
122     return ret % mod;
123 }
124 LL expmod ( LL a, LL b ) {
125     LL res = 1;
126     while ( b ) {
127         if ( b & 1 )  res = res * a % mod;
128         a = a * a % mod;
129         b = b >> 1;
130     }
131     return res % mod;
132 }
133 char s[maxn];
134 int cnt[maxn];
135 int main() {
136    // FIN;
137     scanf ( "%s", s + 1 );
138     n = m = strlen ( s + 1 );
139     LL ans = 0, temp = Manacher ( s + 1, n );
140     n++, m++;
141     for ( int i = 1 ; i < n  ; i++ ) if ( s[i] == 'a' ) a[i] = 1, b[i] = 1;
142     gao();
143     for ( int i = 1 ; i <= 2 * ( n - 1 ) ; i++ ) cnt[i] += res[i];
144     mem ( a, 0 ), mem ( b, 0 ), mem ( res, 0 );
145     for ( int i = 1 ; i < n  ; i++ ) if ( s[i] == 'b' ) a[i] = 1, b[i] = 1;
146     gao();
147     for ( int i = 1 ; i <= 2 * ( n - 1 ) ; i++ ) cnt[i] += res[i];
148     for ( int i = 1 ; i <= 2 * ( n - 1 ) ; i++ )
149         ans = ( ans + expmod ( 2, ( cnt[i] + 1 ) >> 1 ) - 1 ) % mod;
150     printf ( "%lld\n", ( ans - temp + mod ) % mod );
151     return 0;
152 }

 

posted @ 2019-02-26 22:26  Fitz~  阅读(185)  评论(0编辑  收藏  举报