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 tt and 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].
Examples
input
Copy
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
output
Copy
0
1
0
1
input
Copy
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
output
Copy
4
0
3
input
Copy
3 5 2
aaa
baaab
1 3
1 1
output
Copy
0
0#include<bits/stdc++.h>
using namespace std;
int a[1050];
int main()
{
int n,m,k,q,l,r,ans;
memset(a,0,sizeof(a));
string s,t;
cin>>n>>m>>q;
cin>>s>>t;
for(int i=0; i<=n-m; i++)
if(s.substr(i,m)==t)//substr函数其中参数依次是 （ 开始，长度），并返回子串。
a[i]=1;
while(q--)
{
ans=0;
cin>>r>>l;
for(int i=r-1; i<=l-m; i++)
if(a[i])
ans++;
cout<<ans<<endl;
}
return 0;
}
也可以用find函数做
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include<algorithm>
#include <set>
#include <queue>
#include <stack>
#include<vector>
#include<map>
#include<ctime>
#define ll long long
#define INF 0x7fffffff
using namespace std;
struct point
{
int l,r;
}a[1000010];
int main()
{
int n,m,q;
while(cin>>n>>m>>q)
{
string st,t;
cin>>st>>t;
int tot=0;
while(st.find(t)!=string::npos)
{
int x=st.find(t);
int y=x+m-1;
//cout<<x<<" "<<y<<endl;
a[++tot].l=x;
a[tot].r=y;
//cout<<x<<y<<endl;
st[x]='0';
//for(int i=x;i<=y;++i)st[i]='0';这样不对;
//
}
for(int i=1;i<=q;++i)
{
int x,y;
cin>>x>>y;
x-=1;
y-=1;
int ans=0;
for(int j=1;j<=tot;++j)
{
//cout<<x<<" "<<y<<" "<<a[j].l<<" "<<a[j].r<<endl;
if(x<=a[j].l&&y>=a[j].r)ans++;
}
cout<<ans<<endl;
}
}
return 0;
}