【CF245H】【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 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.

Sample test(s)

input

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

output

1
7
3
4
2

注意当s[i] == s[j] 时候不能直接 +1 要判断这个是个回文串才可以，， 因为必需用s[i] 和 s[j] abca 就是个例子

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;

int T;
typedef long long ll;
char s[5002];
ll dp[5002][5002];
int main()
{
		memset(s,0,sizeof(s));
		scanf("%s",s);
		int len =strlen(s);
		for(int i=1;i<=len;i++)
		{
			dp[i][i]  = 1;
			if(i - 1 >= 1)
			dp[i-1][i] = ((s[i-1] == s[i-1-1]) ? 1 : 0);
		}
		for(int j=2;j<=len;j++)
		{
			for(int i=j-2;i>=1;i--)
			{
				if(s[i-1] == s[j-1])
				dp[i][j] += dp[i+1][j-1];
			}
		}
		for(int j=2;j<=len;j++)
		{
			for(int i=j-1;i>=1;i--)
			{
				dp[i][j] = (dp[i][j] + dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1]);
				
			}
		}
		scanf("%d",&T);
		int l,r;
		while(T --)
		{
			scanf("%d%d",&l,&r);
			printf("%I64d\n",dp[l][r]);
		}

    return 0;
}