B. Segment Occurrences
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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
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
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
题意：给定一个字符串 s, 字符串 t. 有q个询问，每次询问输入两个整数 l 和 r, 每次输入 字符串 s 中从 l 位置到 r 位置中 字符串 t 出现的次数(包括 l , r )。 注意 给的 l 和 r 是 第 l 个 到 第 r 个 不是下标。
思路：记录 字符串 s 中 字符串 t 出现的所有位置即可。 查询的时候判断一下 就可以了。
AC代码：
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,m,q;
while(~scanf("%d%d%d",&n,&m,&q)){
string s,t;
string f;
cin >> s >> t;
f = s;
int loc[1050] = {0},plo = 0;
int len = 0;
string::size_type index;
while(1){
index = f.find(t);
if(index != string::npos){
loc[plo ++] = index + len + 1; ///显然, 删除后 字符串会变短，所以要得到在原字符串 s 中的位置 需要加上以 删除的长度
f.erase(0,index + 1); ///每次都要 从下标0 开始删除长度 (index + 1)的字符
len += index + 1; ///如果只删除中间，可能会出现 删除后又组成 t
}else break;
}
int l,r; int ans = 0;
int vis[1050] = {0};
for(int i = 0;i < plo;i ++)
vis[loc[i]] = 1;
for(int i = 1;i <= q;i ++){
scanf("%d%d",&l,&r);
ans = 0;
for(int j = l;j <= r;j ++) ///不仅要判断位置，还要判断长度
if(vis[j] && j + m - 1 <= r) ans ++;
printf("%d\n",ans);
}
}
return 0;
}