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 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
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
题意：给出两个字符串 S 和 T，再给 Q 个区间，求每个区间（针对 S）包含 T 的个数
题解：1>. 暴力 KMP 时间复杂度：Q*（n+m）=1e8 ；能过
2>.预处理，求前缀和。参考：https://blog.csdn.net/PleasantlY1/article/details/81407585
解法1：
#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
#include<set>
#include<map>
#include<math.h>
#include<vector>
#include<bitset>
#include<iostream>
using namespace std;
char s[1050],t[1050];
int nex[1005],len1,len2;
void get(){
int i=0,j=-1;
nex[0]=-1;
while(i<len2){
if(j==-1||t[i]==t[j]){
i++;
j++;
nex[i]=j;
}
else
j=nex[j];
}
}
int sum(int l,int r){
get();
int i=l,j=0,ans=0;
while(i<=r){
if(t[j]==s[i]||j==-1){
i++,j++;
}
else
j=nex[j];
if(j==len2){
j==0;
ans++;
}
}
return ans;
}
int main(){
int n,m,a,b,q;
scanf("%d %d %d",&n,&m,&q);
scanf(" %s",s);
scanf(" %s",t);
len1=strlen(s);
len2=strlen(t);
while(q--){
scanf("%d %d",&a,&b);
printf("%d\n",sum(a-1,b-1));
}
return 0;
}
解法2：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
string s,t;
int main()
{
int n,m,q,l,r;
int ans[1005]={0};
scanf("%d%d%d",&n,&m,&q);
cin>>s>>t;
for(int i=0;i+m<=n;i++)
{
if(s.substr(i,m)==t)
ans[i+1]++;
}
for(int i=1;i<=n;i++)
{
ans[i]+=ans[i-1];
}
while(q--)
{
scanf("%d%d",&l,&r);
r-=m-1;
if(r<0) r=0;
printf("%d\n",max(0,ans[r]-ans[l-1]));
}
return 0;
}