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
Hint
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
题解:
给出两个长为n和m的字符串s和t,给出q次查询,每次给出长度l和r表示查询在字符串s中有多少个t的子串。用substr不断取出上为m的子串,并记录下来,查询的时候直接枚举查询的区间。不了解substr的小伙伴点击这里。某个同学用的find查找,每找到一次记录位置,并将首字母修改,这样可以查找出所有的子串。
代码如下:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<iomanip>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<utility>
#include<list>
#include<algorithm>
#include <ctime>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define swap(a,b) (a=a+b,b=a-b,a=a-b)
#define memset(a,v) memset(a,v,sizeof(a))
#define X (sqrt(5)+1)/2.0
#define maxn 320007
#define N 200005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define mod 1000000009
#define e 2.718281828459045
#define eps 1.0e-8
#define ll long long
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)
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;
}