Codeforces 17E Palisection - Manacher

 1 /**
 2  * Codeforces
 3  * Problem#17E
 4  * Accepted
 5  * Time: 186ms
 6  * Memory: 37200k
 7  */
 8 #include <bits/stdc++.h>
 9 using namespace std;
10 typedef bool boolean;
11 
12 const int N = 2e6 + 5, M = 51123987, inv2 = (M + 1) >> 1;
13 
14 int add(int a, int b) {
15     return ((a += b) >= M) ? (a - M) : (a);
16 }
17 
18 int sub(int a, int b){
19     return ((a -= b) < 0) ? (a + M) : (a);
20 }
21 
22 int n, m;
23 char s[N], str[N << 1];
24 int mxr[N << 1];
25 int f[N];
26 
27 inline void init() {
28     scanf("%d", &n);
29     scanf("%s", s + 1);
30 }
31 
32 void Manacher() {
33     int R = 0, cen = 0;
34     str[0] = ‘+‘;
35     for (int i = 1; i <= n; i++) {
36         str[++m] = s[i];
37         if (i != n)
38             str[++m] = ‘#‘;
39     }
40     str[++m] = ‘-‘;
41     
42     for (int i = 1; i < m; i++) {
43         if (i < R)
44             mxr[i] = min(R - i, mxr[(cen << 1) - i]);
45         while (str[i - mxr[i]] == str[i + mxr[i]])
46             mxr[i]++;
47         if (i + mxr[i] > R)
48             R = i + mxr[i], cen = i;
49     }
50 }
51 
52 inline void solve() {
53     Manacher();
54     for (int i = 1; i < m; i++)
55         f[(i >> 1) + 1]++, f[((i + mxr[i]) >> 1) + 1]--;
56     for (int i = 2; i <= n; i++)
57         f[i] = add(f[i], f[i - 1]);
58     for (int i = 2; i <= n; i++)
59         f[i] = add(f[i], f[i - 1]);
60     int res = sub((f[n] * 1ll * f[n]) % M, f[n]) * 1ll * inv2 % M;
61     for (int i = 2; i <= n; i++)
62         f[i] = add(f[i], f[i - 1]);
63     f[0] = 0, f[n + 1] = 0;
64     
65     for (int i = 1; i <= n; i++) {
66         res = sub(res, sub(f[i - 1], f[max(i - ((mxr[(i - 1) << 1 | 1] + 1) >> 1) - 1, 0)]));
67         if (i < n)
68             res = sub(res, sub(f[i - 1], f[max(i - (mxr[i << 1] >> 1) - 1, 0)]));
69     }
70     printf("%d\n", res);
71 }
72 
73 int main() {
74     init();
75     solve();
76     return 0;
77 }