Queries for Number of Palindromes （区间DP）

[b]Queries for Number of Palindromes[/b]

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 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 qlines 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».

【题意】给你一个字符串，Q次询问，每次给出一个区间问这个区间内有多少个回文字符串。

【分析】ispal[i][j]表示字符串从 i 到 j 这一段是否为回文，若是，则为1。不难得到初始化时ispal[i][i] = 1。那么，对于长度为len的字符串，判断它是否为回文，则有ispal[i][i+len-1] = ispal[i+1][i+len-2]&&str[i]==str[i+len-1];同时ispal[i][i-1]要也要初始化为1，因为当len=2时，ispal[i+1][i+len-2] = ispal[i+1][i]，需要拿来计算。

dp[i][j]表示从 i 到 j 包含的回文串的个数，同样的，初始化dp[i][i] = 1。计数时有：
dp[i][i+len-1] = dp[i][i+len-2]+dp[i+1][i+len-1]-dp[i+1][i+len-2]+ispal[i][i+len-1];然后O(n^2)把每个区间的结果计算出来，对于询问s到e，直接输出dp[s][e]即可。

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 5e3+50;;
const int M = 160009;
const int mod = 1e9+7;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n;
char str
;
int ispal

,dp

;
int main(){
scanf("%s",str+1);
int len=strlen(str+1);
for(int i=1;i<=len;i++)ispal[i][i]=ispal[i][i-1]=dp[i][i]=1;
for(int l=2;l<=len;l++){
for(int i=1;i+l-1<=len;i++){
ispal[i][i+l-1]=ispal[i+1][i+l-2]&(str[i]==str[i+l-1]);
}
}
for(int l=2;l<=len;l++){
for(int i=1;i+l-1<=len;i++){
dp[i][i+l-1]=dp[i][i+l-2]+dp[i+1][i+l-1]-dp[i+1][i+l-2]+ispal[i][i+l-1];
}
}
scanf("%d",&n);
while(n--){
int l,r;
scanf("%d%d",&l,&r);
printf("%d\n",dp[l][r]);
}
return 0;
}