CodeForces - 245H Queries for Number of Palindromes

You’ve got a string s = s1s2… s|s| of length |s|, consisting of lowercase English letters. There also are q queries, each query is described by two integers li, ri (1 ≤ li ≤ ri ≤ |s|). The answer to the query is the number of substrings of string s[li… ri], which are palindromes.

String s[l… r] = slsl + 1… sr (1 ≤ l ≤ r ≤ |s|) is a substring of string s = s1s2… s|s|.

String t is called a palindrome, if it reads the same from left to right and from right to left. Formally, if t = t1t2… t|t| = t|t|t|t| - 1… t1.

Input
The first line contains string s (1 ≤ |s| ≤ 5000). The second line contains a single integer q (1 ≤ q ≤ 106) — the number of queries. Next q lines contain the queries. The i-th of these lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the description of the i-th query.
It is guaranteed that the given string consists only of lowercase English letters.

Output
Print q integers — the answers to the queries. Print the answers in the order, in which the queries are given in the input. Separate the printed numbers by whitespaces.

Examples

input
caaaba
5
1 1
1 4
2 3
4 6

output
1
7
3
4
2

Note
Consider the fourth query in the first test case. String s[4… 6] =«aba». Its palindrome substrings are: «a», «b», «a», «aba».

題意就給你一個串，n次詢問，每次詢問一個區間內有多少個迴文串。
區間型DP

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int SZ = 5010;

char S[SZ];
int get[SZ][SZ];
bool ha[SZ][SZ];

int main()
{
    int n, s, t;
    scanf("%s", S + 1);
    int len = strlen(S + 1);
    for(int i = 1; i <= len; i++)
        ha[i][i] = ha[i + 1][i] = get[i][i] = 1;
    for(int k = 2; k <= len; k++)
        for(int j, i = 1; i + k - 1 <= len; i++)
        {
            j = i + k - 1;
            ha[i][j] = (S[i] == S[j] && ha[i + 1][j - 1]);
            get[i][j] = get[i + 1][j] + get[i][j - 1] - get[i + 1][j - 1] + ha[i][j];
        }
    scanf("%d", &n);
    while(n--)
    {
        scanf("%d%d", &s, &t);
        printf("%d\n", get[s][t]);
    }
    return 0;
}