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
caaaba
5
1 1
1 4
2 3
4 6
4 5
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:求一个区间内回文串的数量
思路:要利用一个判断l~r区间的回文数组来判断这个区间是否是回文串是则为1不是则为0 然后定义一个dp数组来记录l~r区间内回文串的数量
代码:
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<queue> #include<stack> #include<set> #include<vector> #include<map> #include<cmath> const int maxn=1e5+5; typedef long long ll; using namespace std; int hw[5005][5005]; int dp[5005][5005]; char str[5005]; int main() { scanf("%s",str+1); int len=strlen(str+1); for(int t=1;t<=len;t++) { hw[t][t]=1; hw[t][t-1]=1; dp[t][t]=1; } for(int d=1;d<=len;d++) { for(int l=1;l+d<=len;l++) { int r=l+d; if(str[l]==str[r]&&hw[l+1][r-1]) { hw[l][r]=1; } } } for(int d=1;d<=len;d++) { for(int l=1;l+d<=len;l++) { int r=l+d; dp[l][r]=dp[l+1][r]+dp[l][r-1]-dp[l+1][r-1]+hw[l][r]; } } int q; cin>>q; int l,r; while(q--) { scanf("%d%d",&l,&r); printf("%d\n",dp[l][r]); } return 0; }