K - Queries for Number of Palindromes(区间dp+容斥)

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.

Examples

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

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».

题意：求一个区间内回文串的数量

思路：要利用一个判断l~r区间的回文数组来判断这个区间是否是回文串是则为1不是则为0 然后定义一个dp数组来记录l~r区间内回文串的数量

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
int hw[5005][5005];
int dp[5005][5005];
char str[5005];
int main()
{
   scanf("%s",str+1);
   int len=strlen(str+1);
   for(int t=1;t<=len;t++)
   {
       hw[t][t]=1;
       hw[t][t-1]=1;
       dp[t][t]=1;
   }
   for(int d=1;d<=len;d++)
   {
       for(int l=1;l+d<=len;l++)
       {
           int r=l+d;
           if(str[l]==str[r]&&hw[l+1][r-1])
           {
               hw[l][r]=1;
        }
    }
   }
   for(int d=1;d<=len;d++)
   {
       for(int l=1;l+d<=len;l++)
       {
           int r=l+d;
           dp[l][r]=dp[l+1][r]+dp[l][r-1]-dp[l+1][r-1]+hw[l][r];
    }
   }
   int q;
   cin>>q;
   int l,r;
   while(q--)
   {
       scanf("%d%d",&l,&r);
       printf("%d\n",dp[l][r]);
   }
   return 0; 
}