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
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».
题意就给你一个串，n次询问，每次询问一个区间内有多少个回文串。 区间型DP
#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int SZ = 5010; char S[SZ]; int get[SZ][SZ]; bool ha[SZ][SZ]; int main() { int n, s, t; scanf("%s", S + 1); int len = strlen(S + 1); for(int i = 1; i <= len; i++) ha[i][i] = ha[i + 1][i] = get[i][i] = 1; for(int k = 2; k <= len; k++) for(int j, i = 1; i + k - 1 <= len; i++) { j = i + k - 1; ha[i][j] = (S[i] == S[j] && ha[i + 1][j - 1]); get[i][j] = get[i + 1][j] + get[i][j - 1] - get[i + 1][j - 1] + ha[i][j]; } scanf("%d", &n); while(n--) { scanf("%d%d", &s, &t); printf("%d\n", get[s][t]); } return 0; }转载于:https://www.cnblogs.com/Loi-Vampire/p/6017047.html
