CodeForces - 1016B
You are given two strings s s and t t, both consisting only of lowercase Latin letters.
The substring s [ l . . r] s[l..r] is the string which is obtained by taking characters s l , s l + 1 , … , sr sl,sl+1,…,sr without changing the order.
Each of the occurrences of string a a in a string b b is a position i i ( 1 ≤ i ≤ | b | − | a | +1 1≤i≤|b|−|a|+1) such that b [ i . . i + | a | − 1 ] =a b[i..i+|a|−1]=a ( | a | |a| is the length of string a a).
You are asked q q queries: for the i i-th query you are required to calculate the number of occurrences of string t t in a substring s [ l i . . r i] s[li..ri].
Input
The first line contains three integer numbers n n, m m and q q ( 1 ≤ n , m ≤ 103 1≤n,m≤103, 1 ≤ q ≤ 105 1≤q≤105) — the length of string s s, the length of string t t and the number of queries, respectively.
The second line is a string s s ( | s | =n |s|=n), consisting only of lowercase Latin letters.
The third line is a string t t ( | t | =m |t|=m), consisting only of lowercase Latin letters.
Each of the next q q lines contains two integer numbers li li and ri ri ( 1 ≤ l i ≤ r i ≤n 1≤li≤ri≤n) — the arguments for the i i-th query.
Output
Print q q lines — the i i-th line should contain the answer to the i i-th query, that is the number of occurrences of string t t in a substring s [ l i . . r i] s[li..ri].
Examples
Input
10 3 4 codeforces for 1 3 3 10 5 6 5 7
Output
Input
15 2 3 abacabadabacaba ba 1 15 3 4 2 14
Output
Input
3 5 2 aaa baaab 1 3 1 1
Output
Note
In the first example the queries are substrings: " cod", " deforces", " fo" and "for", respectively.
剛開始不會做，是因為只想到了用字首和，沒想到應該怎麼去用，根據題意，只有當string t 全部在在所查區間的時候，答案才加一，
所以說如果從查詢區間的頭部向後遍歷的話，不但要知道從哪個字母開始，還要判斷字串t何時結束，字首和不太好維護。
因此，我們從查詢區間的尾部向前遍歷，只要找到字串t結束的地方，只需要判斷字串t的開頭是否在查詢區間之內即可，這樣的字首和容易維護。
1 #include<cstdio>
2 #include<cstdlib>
3 #include<cstring>
4 #include<string>
5 #include<cmath>
6 #include<algorithm>
7 #include<queue>
8 #include<stack>
9 #include<deque>
10 #include<map>
11 #include<iostream>
12 using namespace std;
13 typedef long longLL;
14 const double pi=acos(-1.0);
15 const double e=exp(1);
16 const int N = 10;
17
18 char con[1009],sub[1009];
19 int check[1009];
20 int main()
21 {
22int i,p,j,n,m,q;
23int flag,a,b,ans;
24scanf("%d%d%d",&n,&m,&q);
25scanf(" %s",con+1);
26scanf(" %s",sub+1);
27
28for(i=1;i<=n-m+1;i++)
29{
30p=1;
31flag=0;
32for(j=i;j<=i+m;j++)
33{
34if(p>m)
35{
36flag=1;
37break;
38}
39if(con[j]!=sub[p])
40break;
41p++;
42}
43if(flag==1)
44check[j-1]=1;
45}
46for(i=1;i<=q;i++)
47{
48ans=0;
49scanf("%d%d",&a,&b);
50for(j=a;j<=b;j++)
51{
52if(check[j]==1&&j-(m-1)>=a)
53ans++;
54}
55printf("%d\n",ans);
56}
57return 0;
58 }
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