10算法------补题，注意时间复杂度的问题_a. good pairs time limit per test1 second memory l

题的链接：
https://codeforces.com/group/mey3UXMrvB/contest/313754/problem/B

B. Segment Occurrences
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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].

Examples
inputCopy
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
outputCopy
0
1
0
1
inputCopy
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
outputCopy
4
0
3
inputCopy
3 5 2
aaa
baaab
1 3
1 1
outputCopy
0
0
Note
In the first example the queries are substrings: “cod”, “deforces”, “fo” and “for”, respectively.

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
void solve()
{
    ll n,m,q;
    cin>>n>>m>>q;
    string s,t;
    cin>>s>>t;
    ll ans[n+1]={0};
    ll an=0;
    for(int i=0;i<=n-m;i++)
    {
        if(s.substr(i,m)==t) ans[i]=1;//注意substr的复杂度为o（n）
    }
    for(ll i=0;i<q;i++)
{
        ll l,r;
        cin>>l>>r;
        for(int i=l-1;i<=r-m;i++)
        {
            if(ans[i]) an++;
        }
        cout<<an<<endl;
        an=0;
    }
}
 
 
 
int main()
{
    solve();
    return 0;
}

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