245H Queries for Number of Palindromes

H. Queries for Number of Palindromes

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 l_i, r_i (1 ≤ l_i ≤ r_i ≤ |s|). The answer to the query is the number of substrings of string s[l_i... r_i], which are palindromes.

String s[l... r] = s_ls_l + 1... s_r (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 l_i, r_i (1 ≤ l_i ≤ r_i ≤ |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.

Sample test(s)

input

caaaba
5
1 1
1 4
2 3
4 6
4 5

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».

　　本题求任意子串包含的回文数，DP问题，状态转移方程：d[i][j]=d[i+1][j]+d[i][j-1]-d[i+1][j-1]+ispalindrome(i,j)，其中ispalindrome(i,j)=ispalindrome(i+1,j-1)&&s[i]==s[j].

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[5010][5010];
bool is[5010][5010];
char s[5010];
int main(){
    int i,len,q,a,b;
    scanf("%s",s);
    len=strlen(s);
    for(i=0;i<len;++i)
        dp[i][i]=is[i][i]=1;
    for(i=2;i<=len;++i){
        for(a=0;a<len+1-i;++a){
            b=a+i-1;
            if((a+1>b-1 || is[a+1][b-1]) && s[a]==s[b])
                is[a][b]=1;
            dp[a][b]=dp[a+1][b]+dp[a][b-1]-dp[a+1][b-1]+is[a][b];
        }
    }
    scanf("%d",&q);
    while(q--){
        scanf("%d%d",&a,&b);
        printf("%d\n",dp[a-1][b-1]);
    }
    return 0;
}