codeforces 17E_牛客博客

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 to z).

Output

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

        Examples       
          input                     Copy          
4
babb
          output                     Copy          
6
          input                     Copy          
2
aa
          output                     Copy          
2
          
         

题意：

求一个字符串中相交回文串的对数。

题解：

先求字符串中回文串的总对数，再减去不相交回文串的对数。

用Manacher求出回文半径，求和就是回文串的个数，平方除2就是回文串总对数。

下面求不相交回文串的对数。

设s[i]表示以i为结束的回文串的个数，s1[i]表示以第j（1<=j<=i）为结束的回文子串的个数和。设当前处理到以第i位为中心的回文子串与左侧回文子串的不相交对数，即

为了快速求得上式，再求前缀和即可。

代码：

#include<bits/stdc++.h>
#define N 2000010
#define LL long long
using namespace std;
char s[N],ss[N<<1];
int r[N<<1],len;LL s1[N];

int Init()
{
    int len = strlen(s);
    ss[0] = '$';ss[1] = '#';
    int j = 2;
    for (int i = 0; i < len; i++)ss[j++] = s[i],ss[j++] = '#';
    ss[j] = '\0';
    return j;
}

void Manacher()
{
    len=Init();
    int p,mx=0;
    for (int i = 1; i < len; i++)
    {
        if (i<mx) r[i]=min(r[2*p-i],mx-i);else r[i] = 1;
        while (ss[i-r[i]]==ss[i+r[i]]) r[i]++;
        if (mx<i+r[i])p=i,mx=i+r[i];
    }
}

int main()
{
    int n;
    while (~scanf("%d%s",&n,s))
    {
        Manacher();
        LL tot=0;
        memset(s1,0,sizeof s1);
        for (int i=1;i<=n*2;i++)
        {
            s1[i/2+(i&1)]++;
            s1[i/2+(i&1)+r[i]/2]--;
            tot=(tot+r[i]/2)%51123987;
        }
        tot=tot*(tot-1)/2%51123987;
        for (int i=2;i<=n;i++) s1[i]+=s1[i-1];
        for (int i=2;i<=n;i++) s1[i]+=s1[i-1];
        for (int i=2;i<=n;i++) s1[i]+=s1[i-1];
        LL ans=0;
        for (int i=2;i<=n*2;i++) ans+=s1[i/2-1]-s1[i/2-r[i]/2-1],ans%=51123987;
        printf("%I64d\n",(tot-ans+51123987)%51123987);
    }
}