51nod 1732 51nod Marriage Institute

51nod In addition to doing OJ, but also carried out a lot of sideline. Marriage introduction is one of them.
For a customer, we can use a string to describe the characteristics of the customer.
Suppose we now have two customers A and B.
The character string for a is: ABCDEFGB's character string is: abcxyz
Then A and B match degrees F (A, B) is the length of the longest public prefix of a and B, that is Len (' abc ') = 3
Because of the recent 51nod budget, the jacket master designed a compression algorithm to conserve memory.
The trait strings for all users are stored in a string s of length n. (n <= 1000) The user's trait is represented by an integer p, which indicates that the user's trait string is s[p...n-1].
Now given the string s, and q query <ai, bi> (AI, Bi is a legitimate user trait integer). Please output the Q query corresponding to the customer matching degree.



Input

Now given the string length n, with the string s. Next is the integer q, which represents the next Q-Query. The following Q line has two integer ai, bi. Represents the matching degree of the query trait for users of AI and bi. 1 <= n <= 10001 <= q <= 10^6 input data all valid.

Output

Each row outputs a user-matching integer.

Input example

12loveornolove53 70 09 13 19 5

Output example

012300

Wizmann(topic Provider)Idea: Directly reverse the string, and then Dp[i][j] represents the first and the longest public suffix of J, the last query directly output DP[N-U][N-V] good, O (n^2) preprocessing, O (1) query.

#include <iostream> #include <algorithm> #include <cstdio> #include <queue> #include <map > #include <vector> #include <cstring> #include <cmath>using namespace std;typedef long Long ll;    const int INF =0x3f3f3f3f;const double pi = ACOs ( -1.0); const int N = 1e3 + 10;int Dp[n][n];char s[n],x[n];int main () {    int n, m;    scanf ("%d", &n);    scanf ("%s", s);    for (int i = 0; i<n; i++) {x[n-i-1] = s[i];    } X[n] = ' + ';    memset (DP, 0, sizeof (DP));                for (int i = 1, i<=n; i++) for (int j = 1; j<=n; j + +) {if (x[i-1] = = X[j-1]) {            DP[I][J] = dp[i-1][j-1] + 1;        } else Dp[i][j] = 0;        } int u, v;        scanf ("%d", &m);            while (m--) {scanf ("%d%d", &u, &v);        printf ("%d\n", Dp[n-u][n-v]); } return 0;}

51nod 1732 51nod Marriage Institute