H. Queries for Number of Palindromes
time limit per test
memory limit per test
input
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
Sample test(s)
Input
caaaba51 11 42 34 64 5
Output
17342
Note
Consider the fourth query in the first test case. String s[4... 6] = «aba». Its palindrome substrings are: «a», «b», «a», «aba».
思路：DP，用个二维布尔数组判断是否回文。在用个二维数组存最优值
初始化，当只有一个字符时是回文，并且最优值是1
状态方程：is_palin[j][j+i]=is_palin[j+1][j+i-1]&s[j]==s[j+i]; dp[j][j+i]=dp[j+1][j+i]+dp[j][j+i-1]-dp[j+1][i+j-1]+is_palin[j][j+i];
#include<iostream>#include<cstring>using namespace std;const int mm=5005;char s[mm];int dp[mm][mm];bool is_palin[mm][mm];int main(){ cin>>s; int len=strlen(s); memset(is_palin,0,sizeof(is_palin)); memset(dp,0,sizeof(dp)); for(int i=0;i<len;i++) { dp[i][i]=1; is_palin[i][i]=true; is_palin[i+1][i]=true; } for(int i=1;i<=len;i++) for(int j=0;j<=len-i;j++) { is_palin[j][j+i]=is_palin[j+1][j+i-1]&s[j]==s[j+i]; dp[j][j+i]=dp[j+1][j+i]+dp[j][j+i-1]-dp[j+1][i+j-1]+is_palin[j][j+i]; } int m,a,b; cin>>m; for(int i=0;i<m;i++) { cin>>a>>b; cout<<dp[a-1][b-1]<<"\n"; }}