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.
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.
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.
caaaba 5 1 1 1 4 2 3 4 6 4 5
1 7 3 4 2
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define SIZE 5001
using namespace std;
int ispal[SIZE][SIZE];
int dp[SIZE][SIZE];
char str[SIZE];
int Q;
int main()
{
while(~scanf("%s",str+1))
{
int len = (int)strlen(str+1);
for(int i=1; i<=len; i++)
{
dp[i][i] = ispal[i][i] = 1;
ispal[i][i-1] = 1;
}
for(int i=2; i<=len; i++)
{
for(int s=1; s+i-1 <=len; s++)
{
ispal[s][s+i-1] = ispal[s+1][s+i-2]&&str[s]==str[s+i-1];
dp[s][s+i-1] = dp[s][s+i-2]+dp[s+1][s+i-1]-dp[s+1][s+i-2]+ispal[s][s+i-1];
}
}
scanf("%d",&Q);
int s,e;
while(Q--)
{
scanf("%d%d",&s,&e);
printf("%d\n",dp[s][e]);
}
}
return 0;
}
本文介绍了一种高效计算字符串中特定区间内回文子串数量的算法,并提供了完整的C++实现代码。该算法通过预处理得到所有可能的回文子串信息,能够在O(1)时间内回答任意区间的回文子串数量。
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