【CodeForces - 245H 】Queries for Number of Palindromes （带容斥的区间dp）

题干：

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
4 5

Output

Note

Consider the fourth query in the first test case. String s[4... 6] = «aba». Its palindrome substrings are: «a», «b», «a», «aba».

题目大意：

现有一个字符串 s = s1s2... s|s| of length |s|, 由小写字符组成. 现在有 q 次查询, 每次查询给两个整数 li, ri (1 ≤ li ≤ ri ≤ |s|). 每次查询你的程序要给出此字符串的子串 s[li... ri]有多少个回文串.

解题报告：

显然是区间dp，，刚开始写搓了，还是区间dp不太熟练。。

正经题解：先处理处长度为2的来，并且预处理出任意 l r 子串是否是回文串。然后从len=3开始跑区间dp就行了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 5000 + 5;
char s[MAX];
int dp[MAX][MAX];
bool is[MAX][MAX];
int l,r;
int main()
{cin>>(s+1);int n = strlen(s+1);for(int i = 1; i<=n; i++) dp[i][i] = 1,is[i][i]=1;for(int i = 1; i<=n-2+1; i++) {dp[i][i+1]=2;  if(s[i] == s[i+1]) dp[i][i+1]++,is[i][i+1]=1;}for(int len = 3; len<=n; len++) {for(int l = 1; l<=n-len+1; l++) {int r = l+len-1;if(s[l] == s[r]) is[l][r] = is[l+1][r-1];}}for(int len = 3; len<=n; len++) {for(int l = 1; l<=n-len+1; l++) {int r = l+len-1;dp[l][r] = dp[l+1][r] + dp[l][r-1] - dp[l+1][r-1];if(is[l][r]) dp[l][r]++;}}int q;cin>>q;while(q--) {scanf("%d%d",&l,&r);printf("%d\n",dp[l][r]);}return 0 ;}

另一个AC代码：（好像我写的那个跑1600ms，，这个代码跑200+ms））本来以为可能是用了按位与，，但是发现这样跑出来也不是很快啊1700+ms。。。可能是数据加强了？

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 5000 + 5;
char s[MAX];
int dp[MAX][MAX];
bool is[MAX][MAX];
int l,r;
int main()
{cin>>(s+1);int n = strlen(s+1);for(int i = 1; i<=n; i++) dp[i][i] = 1,is[i][i]=1,is[i+1][i]=1;
//  for(int i = 1; i<=n-2+1; i++) {
//      dp[i][i+1]=2;
//      if(s[i] == s[i+1]) dp[i][i+1]++,is[i][i+1]=1;
//  }for(int len = 1; len<=n; len++) {for(int j = 1; j<=n; j++) {is[j][j+len] = is[j+1][j+len-1] & (s[j] == s[j+len]);dp[j][j+len] = dp[j][j+len-1] + dp[j+1][j+len] - dp[j+1][j+len-1] + is[j][j+len];}}int q;cin>>q;while(q--) {scanf("%d%d",&l,&r);printf("%d\n",dp[l][r]);}return 0 ;}

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【CodeForces - 245H 】Queries for Number of Palindromes （带容斥的区间dp）

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