Codeforces Beta Round #17 E. Palisection(回文树)

E. Palisection

time limit per test

2 seconds

memory limit per test

128 megabytes

input

standard input

output

standard output

In an English class Nick had nothing to do at all, and remembered about wonderful strings called palindromes. We should remind you that a string is called a palindrome if it can be read the same way both from left to right and from right to left. Here are examples of such strings: «eye», «pop», «level», «aba», «deed», «racecar», «rotor», «madam».

Nick started to look carefully for all palindromes in the text that they were reading in the class. For each occurrence of each palindrome in the text he wrote a pair — the position of the beginning and the position of the ending of this occurrence in the text. Nick called each occurrence of each palindrome he found in the text subpalindrome. When he found all the subpalindromes, he decided to find out how many different pairs among these subpalindromes cross. Two subpalindromes cross if they cover common positions in the text. No palindrome can cross itself.

Let's look at the actions, performed by Nick, by the example of text «babb». At first he wrote out all subpalindromes:

• « b» —  1..1 • « bab» —  1..3 • « a» —  2..2 • « b» —  3..3 • « bb» —  3..4 • « b» —  4..4

Then Nick counted the amount of different pairs among these subpalindromes that cross. These pairs were six:

1.  1..1 cross with  1..3 2.  1..3 cross with  2..2 3.  1..3 cross with  3..3 4.  1..3 cross with  3..4 5.  3..3 cross with  3..4 6.  3..4 cross with  4..4

Since it's very exhausting to perform all the described actions manually, Nick asked you to help him and write a program that can find out the amount of different subpalindrome pairs that cross. Two subpalindrome pairs are regarded as different if one of the pairs contains a subpalindrome that the other does not.

Input

The first input line contains integer n (1 ≤ n ≤ 2·106) — length of the text. The following line contains n lower-case Latin letters (from a toz).

Output

In the only line output the amount of different pairs of two subpalindromes that cross each other. Output the answer modulo 51123987.

Sample test(s)

input

4
babb

output

6

input

2
aa

output

2

给出一个长度为ｎ的字符串，所有回文串集合为Ｓ，问从Ｓ中任取两个有公共部分的有多少种方案，直接求相交的不好求，转成总的减不相交的，本题要用邻接表否则会ＭＬＥ

#include 
using namespace std;
#define fi first
#define se second
typedef long long ll;
typedef vector >vii;
const int N = 26;
const int MAXN = 2e6 +20;
const int mod = 51123987;
struct Palindromic_Tree {
    vii next[MAXN];
    int fail[MAXN],len[MAXN],S[MAXN],num[MAXN];
    int last,n,p;
    int newnode(int l)
    {
        next[p].clear();
        num[p] = 0;
        len[p] = l;
        return p++;
    }
    int find(int u,int c)
    {
        vii & x = next[u];
        int sz = x.size();
        for(int i = 0; i < sz; ++i) {
            if(x[i].fi == c) return x[i].se;
        }
        return 0;
    }
    void init()
    {
        last = n = p = 0;
        S[0] = -1;
        newnode(0);
        newnode(-1);
        fail[0] = 1;
        fail[1] = 0;
    }
    int get_fail(int x)
    {

        while(S[n-len[x]-1] != S[n]) {
            x  = fail[x];
        }
        return x;
    }
    int add(int c)
    {
        c -= 'a';
        S[++n] = c;
        int cur = get_fail(last);
        int x = find(cur,c);
        if(x==0) {
            int now = newnode(len[cur]+2);
            x = now;
            fail[now] = find(get_fail(fail[cur]),c);
            next[cur].push_back(make_pair(c,now));
            num[now] = num[fail[now]] + 1;
        }
        last = x;
        return num[last];
    }
} ;
Palindromic_Tree T;
char s[MAXN];
ll L[MAXN],R[MAXN];
int main()
{
    int n;
    while(scanf("%d%s",&n,s+1)==2) {
        memset(L,0,sizeof(*L)*(n+10));
        memset(R,0,sizeof(*R)*(n+10));
        T.init();
        for(int i = 1; i <= n; ++i) {
            R[i] = T.add(s[i]);
        }
        T.init();
        for(int i = n; i >= 1; --i) {
            L[i] = T.add(s[i]);
        }
        for(int i = n-1; i >= 1; --i) L[i] =(L[i] + L[i+1])%mod;
        ll tot = L[1];
        tot %= mod;
        ll ans = tot*(tot-1)/2;
        for(int i = 1; i <= n; ++i) {
            ans = (ans + mod-(R[i]*L[i+1]%mod)) % mod;
        }
        cout<