回文串的组合 DP

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You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.

In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b],s[x... y] are palindromes.

A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.

A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".

Input

The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.

Output

Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.

Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.

Examples

input
aa

output
1

input
aaa

output
5

input
abacaba

output
36

题目大意：

回文串的组合

http://codeforces.com/contest/159/problem/D

思路:
虽然表示题解说的挺详细了，但是敲起来的时候还是遇到了各种问题

首先我们定义p[j][i]代表j到i是否是会问，然后用dp[i]来保存j<i的回文总数就好了

#include
#include
#include
#include
#include
#include
#include

using namespace std;

typedef long long ll;
typedef pair  pii;
const int maxn = 2000 + 5;

int p[maxn][maxn];
char ch[maxn];
int n;
ll dp[maxn];
ll sum[maxn];

int main(){
    scanf("%s", ch);
    int lo = 0;
    for (int i = 0; ch[i] != '\0'; i++){
        lo++;
    }
    for (int i = 0; i < lo; i++){
        p[i][i] = 1;
        for (int l = i - 1, r = i + 1; l >= 0 && r < lo; l--, r++){
            if (ch[l] != ch[r]) break;
            p[l][r] = 1;
        }
        for (int l = i, r = i + 1; l >= 0 && r < lo; l--, r++){
            if (ch[l] != ch[r]) break;
            p[l][r] = 1;
        }
    }
    for (int i = 0; i < lo; i++){
        for (int j = 0; j <= i; j++){
            sum[i] += p[j][i];
        }
    }
    for (int i = 1; i < lo; i++){
        sum[i] += sum[i-1];
    }
    ll ans = 0;
    for (int i = 0; i < lo; i++){
        for (int j = 0; j <= i; j++){
            if (p[j][i]) ans += sum[j - 1];
        }
    }
    printf("%I64d\n", ans);
    return 0;
}