You are given two strings s and t , both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters sl,sl+1,…,sr without changing the order.
Each of the occurrences of string a in a string b is a position i (1≤i≤|b|−|a|+1 ) such that b[i..i+|a|−1]=a (|a| is the length of string a ).
You are asked q queries: for the i -th query you are required to calculate the number of occurrences of string t in a substring s[li..ri] .
The first line contains three integer numbers n , m and q (1≤n,m≤103 , 1≤q≤105 ) — the length of string s , the length of string t and the number of queries, respectively.
The second line is a string s (|s|=n ), consisting only of lowercase Latin letters.
The third line is a string t (|t|=m ), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers li and ri (1≤li≤ri≤n ) — the arguments for the i -th query.
Print q lines — the i -th line should contain the answer to the i -th query, that is the number of occurrences of string t in a substring s[li..ri] .
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
0
1
0
1
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
4
0
3
3 5 2
aaa
baaab
1 3
1 1
0
0
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
题目大意：就是给两个字符串s t，然后q次查询，给出 [l, r], 问t出现的次数。
刚开始做这道题感觉就是瞎写，没有好好思考，下面给出官方的思路：首先看一下单纯的做法。q次查询，每次从 i 属于 [l, r-m+1] 然后遍历，看是否和t一样。时间复杂度(q*m*n).
注意到t只能从s的n个位置开始，我们可以预处理t出现的位置，然后前缀和维护出现次数，这样的话，每次查询都是O(1).
1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <string.h> 5 #include <stdlib.h> 6 #include <string> 7 #include <vector> 8 #include <set> 9 #include <map> 10 #include <queue> 11 #include <algorithm> 12 #include <sstream> 13 #include <stack> 14 using namespace std; 15 typedef long long ll; 16 const int inf = 0x3f3f3f3f; 17 18 const int N = 1000 + 7; 19 int pre[N]; 20 21 int main() { 22 //freopen("in.txt", "r", stdin); 23 int n, m, q; 24 scanf("%d%d%d", &n, &m, &q); 25 string s, t; 26 cin >> s >> t; 27 for(int i = 0; i < n - m + 1; i++) {//从s中找t开始的位置 28 bool flag = true; 29 for(int j = 0; j < m; j++) { 30 if(s[i + j] != t[j]) 31 flag = false; 32 } 33 pre[i+1] = pre[i] + flag;//前缀和 34 } 35 for(int i = max(0, n - m + 1); i < n; i++)//上面终止条件，n-m+1 后面的pre还没有值 36 pre[i+1] = pre[i]; 37 for(int i = 0; i < q; i++) { 38 int l, r; 39 scanf("%d%d", &l, &r); 40 l--, r -= m - 1;//r -= m-1 变成起始位置（本次次数），l-- 变成上次出现次数 41 printf("%d ", l <= r ? pre[r] - pre[l] : 0); 42 } 43 }