CodeForces

先上题目：

H. Queries for Number of Palindromes

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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 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.

Sample test(s)

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：给你一个字符串，q个查询，问你某个区间里面有多少个回文串。

　　区间dp,需要三个数组: s[] 字符串；f[i][j] [i,j]是不是一个回文串；dp[i][j] [i,j]包含了多少个字符串。

　　初始化：

　　f[i][i]=1

　　f[i][i-1]=1           //在判断偶数长度的回文串的时候需要使用

　　dp[i][i]=1

　　状态转移方程：   dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+f[i][j]

　　为什么可以这样列方程？这需要说一下对字符串的操作：我们先判断短的字符串，然后再增长判断的长度，这样dp[i+1][j]、dp[i][j-1]和dp[i+1][j-1]在对dp[i][j]操作之前就已经处理完保存了处理以后的数据了。

上代码：

#include <cstdio>
#include <cstring>
#define MAX 5002
using namespace std;

char s[MAX];

/*
    dp[i][j]: [i,j]中包含了多少个回文串
    f[i][j]:  [i,j]是否是回文串
*/
int dp[MAX][MAX];
int f[MAX][MAX];

int main()
{
    int q,a,b,l;
    //freopen("data.txt","r",stdin);
    while(scanf("%s",s+1)!=EOF){
        memset(dp,0,sizeof(dp));
        memset(f,0,sizeof(f));
        l=strlen(s+1);
        for(int i=1;i<=l;i++){
                /*初始化:
                         f[i][i]=1   单个字母是一个回文串
                         f[i][i-1]=1 考虑偶数个字母是一个回文串的时候的情况(状态转移的方向决定的)
                         dp[i][i]=1   单个字母的时候[i,i]有一个回文串
                */
                f[i][i]=1;
                f[i][i-1]=1;
                dp[i][i]=1;
        }
        for(int le=2;le<=l;le++){
            for(int i=1,j=le;j<=l;i++,j++){
                if(s[i]==s[j]){
                   /*s[i]==s[j] &&  [i+1,j-1]也是回文串*/
                   f[i][j]=f[i+1][j-1];
                }
                dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+f[i][j];
            }
        }
        scanf("%d",&q);
        while(q--){
            scanf("%d %d",&a,&b);
            printf("%d
",dp[a][b]);
        }

    }
    return 0;
}