Description
You are given two strings ss and tt, both consisting only of lowercase Latin letters.
The substring s[l..r]s[l..r] is the string which is obtained by taking characters sl,sl+1,…,srsl,sl+1,…,sr without changing the order.
Each of the occurrences of string aa in a string bb is a position ii (1≤i≤|b|−|a|+11≤i≤|b|−|a|+1) such that b[i..i+|a|−1]=ab[i..i+|a|−1]=a (|a||a| is the length of string aa).
You are asked qq queries: for the ii-th query you are required to calculate the number of occurrences of string tt in a substring s[li..ri]s[li..ri].
Input
The first line contains three integer numbers nn, mm and qq (1≤n,m≤1031≤n,m≤103, 1≤q≤1051≤q≤105) — the length of string ss, the length of string ttand the number of queries, respectively.
The second line is a string ss (|s|=n|s|=n), consisting only of lowercase Latin letters.
The third line is a string tt (|t|=m|t|=m), consisting only of lowercase Latin letters.
Each of the next qq lines contains two integer numbers lili and riri (1≤li≤ri≤n1≤li≤ri≤n) — the arguments for the ii-th query.
Output
Print qq lines — the ii-th line should contain the answer to the ii-th query, that is the number of occurrences of string tt in a substring s[li..ri]s[li..ri].
Sample Input
Input
10 3 4 codeforces for 1 3 3 10 5 6 5 7
Output
0 1 0 1
Input
15 2 3 abacabadabacaba ba 1 15 3 4 2 14
Output
4 0 3
Input
3 5 2 aaa baaab 1 3 1 1
Output
0 0
求下边字符串再上边字符串中出现的次数。emmm比赛没想到好方法，不出所料的TLE，学会一个find().
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<iomanip>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
using namespace std;
typedef long long int LL;
const int MAXN=1e5+10;
map<int,int> mp;
int main()
{
int n,m,q,x,y;
string a,b;
scanf("%d %d %d",&n,&m,&q);
cin>>a>>b;
while(1)
{
int t=a.find(b);
if(t!=-1)
{
mp[t]++;
a[t]='3';
}
else
break;
}
while(q--)
{
int s;
scanf("%d %d",&x,&y);
x--;
y--;
s=0;
map<int,int>::iterator it;
for(it=mp.begin(); it!=mp.end(); it++)
{
if(it->first>=x && it->first<=y-m+1)
s++;
}
printf("%d\n",s);
}
return 0;
}