Codeforces 245H H Queries for Number of Palindromes（DP）

H. Queries for Number of Palindromes

time limit per test

memory limit per test

input

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

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».

思路：DP，用个二维布尔数组判断是否回文。在用个二维数组存最优值

         初始化，当只有一个字符时是回文，并且最优值是1

         状态方程：is_palin[j][j+i]=is_palin[j+1][j+i-1]&s[j]==s[j+i];
    dp[j][j+i]=dp[j+1][j+i]+dp[j][j+i-1]-dp[j+1][i+j-1]+is_palin[j][j+i];

#include<iostream>
#include<cstring>
using namespace std;
const int mm=5005;
char s[mm];
int dp[mm][mm];
bool is_palin[mm][mm];
int main()
{
  cin>>s;
  int len=strlen(s);
  memset(is_palin,0,sizeof(is_palin));
  memset(dp,0,sizeof(dp));
  for(int i=0;i<len;i++)
  {
    dp[i][i]=1;
    is_palin[i][i]=true;
    is_palin[i+1][i]=true;
  }
  for(int i=1;i<=len;i++)
  for(int j=0;j<=len-i;j++)
  {
    is_palin[j][j+i]=is_palin[j+1][j+i-1]&s[j]==s[j+i];
    dp[j][j+i]=dp[j+1][j+i]+dp[j][j+i-1]-dp[j+1][i+j-1]+is_palin[j][j+i];
  }
  int m,a,b;
  cin>>m;
  for(int i=0;i<m;i++)
  {
    cin>>a>>b;
    cout<<dp[a-1][b-1]<<"\n";
  }
}