题干：
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 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».
题目大意：
现有一个字符串 s = s1s2... s|s| of length |s|, 由小写字符组成. 现在有 q 次查询, 每次查询给两个整数 li, ri (1 ≤ li ≤ ri ≤ |s|). 每次查询你的程序要给出此字符串的子串 s[li... ri]有多少个回文串.
解题报告：
显然是区间dp，，刚开始写搓了，还是区间dp不太熟练。。
正经题解：先处理处长度为2的来，并且预处理出任意 l r 子串是否是回文串。然后从len=3开始跑区间dp就行了。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 5000 + 5;
char s[MAX];
int dp[MAX][MAX];
bool is[MAX][MAX];
int l,r;
int main()
{
cin>>(s+1);
int n = strlen(s+1);
for(int i = 1; i<=n; i++) dp[i][i] = 1,is[i][i]=1;
for(int i = 1; i<=n-2+1; i++) {
dp[i][i+1]=2;
if(s[i] == s[i+1]) dp[i][i+1]++,is[i][i+1]=1;
}
for(int len = 3; len<=n; len++) {
for(int l = 1; l<=n-len+1; l++) {
int r = l+len-1;
if(s[l] == s[r]) is[l][r] = is[l+1][r-1];
}
}
for(int len = 3; len<=n; len++) {
for(int l = 1; l<=n-len+1; l++) {
int r = l+len-1;
dp[l][r] = dp[l+1][r] + dp[l][r-1] - dp[l+1][r-1];
if(is[l][r]) dp[l][r]++;
}
}
int q;
cin>>q;
while(q--) {
scanf("%d%d",&l,&r);
printf("%d\n",dp[l][r]);
}
return 0 ;
}