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].
Input
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.
Output
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]
input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
ouput
0
1
0
1
input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
ouput
4
0
3
input
3 5 2
aaa
baaab
1 3
1 1
ouput
0
0
题意:找区间内b串的个数，
做法:扫一遍就行
#include<stdio.h>
#include<bits/stdc++.h>
#define inf 1000000007
using namespace std;
string a,b;
int we[10000];
int main()
{
int n,m,q;
while(scanf("%d %d %d",&n,&m,&q)!=EOF){
memset(we,0,sizeof(we));
cin>>a>>b;
if(n>=m){
int s=a.find(b);
while(s!=string::npos){
we[s+m-1]++;
s=a.find(b,s+1);
}
for(int i=1;i<n;i++){
we[i]+=we[i-1];
}
}
int l,r;
while(q--){
scanf("%d %d",&l,&r);
if(m>n){
printf("0\n");
continue;
}
if(r-l<m-1)
printf("0\n");
else
printf("%d\n",we[r-1]-we[l+m-3]);//一定要注意左区间的取法
}
}
return 0;
}