K - Queries for Number of Palindromes(区间dp+容斥)-白红宇

K - Queries for Number of Palindromes(区间dp+容斥)

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发布时间：2019-06-12

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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 l_i, r_i (1 ≤ l_i ≤ r_i ≤ |s|). The answer to the query is the number of substrings of string s[l_i... r_i], which are palindromes.

String s[l... r] = s_ls_l + 1... s_r (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 l_i, r_i (1 ≤ l_i ≤ r_i ≤ |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：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
     
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          #include
          
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             #include
              
               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; }

转载于:https://www.cnblogs.com/Staceyacm/p/11240949.html

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