xtu summer individual-4 D

Martian Strings

Time Limit: 2000ms

Memory Limit: 262144KB

This problem will be judged on CodeForces. Original ID: 149E
64-bit integer IO format: %I64d      Java class name: (Any)

 

During the study of the Martians Petya clearly understood that the Martians are absolutely lazy. They like to sleep and don't like to wake up.

Imagine a Martian who has exactly n eyes located in a row and numbered from the left to the right from 1 to n. When a Martian sleeps, he puts a patch on each eye (so that the Martian morning doesn't wake him up). The inner side of each patch has an uppercase Latin letter. So, when a Martian wakes up and opens all his eyes he sees a string sconsisting of uppercase Latin letters. The string's length is n.

"Ding dong!" — the alarm goes off. A Martian has already woken up but he hasn't opened any of his eyes. He feels that today is going to be a hard day, so he wants to open his eyes and see something good. The Martian considers only m Martian words beautiful. Besides, it is hard for him to open all eyes at once so early in the morning. So he opens two non-overlapping segments of consecutive eyes. More formally, the Martian chooses four numbers a, b, c, d, (1 ≤ a ≤ b < c ≤ d ≤ n) and opens all eyes with numbers i such that a ≤ i ≤ b or c ≤ i ≤ d. After the Martian opens the eyes he needs, he reads all the visible characters from the left to the right and thus, he sees some word.

Let's consider all different words the Martian can see in the morning. Your task is to find out how many beautiful words are among them.

 

Input

The first line contains a non-empty string s consisting of uppercase Latin letters. The strings' length is n (2 ≤ n ≤ 105). The second line contains an integer m (1 ≤ m ≤ 100) — the number of beautiful words. Next m lines contain the beautiful words pi, consisting of uppercase Latin letters. Their length is from 1 to 1000. All beautiful strings are pairwise different.

 

Output

Print the single integer — the number of different beautiful strings the Martian can see this morning.

 

Sample Input

Input

ABCBABA
2
BAAB
ABBA

Output

1

Hint

Let's consider the sample test. There the Martian can get only the second beautiful string if he opens segments of eyes a = 1, b = 2 and c = 4, d = 5 or of he opens segments of eyes a = 1, b = 2 and c = 6, d = 7.

 

Source

Codeforces Round #106 (Div. 2)

 

 

解题：KMP，几天没写KMP，居然犯了个低级错误！题意大概就是给出一个母串，再给出若干字串，要求统计有多少字串可以优母串中的两个字串拼接而成！注意顺序，样例说得很清楚了，左边的一定要在左边。假设a<b<c<d，abcd这样拼接是可以的，而cdab是禁止的。注意母串逆转过去后，在下次操作前，要逆转回去。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#define LL long long
#define INF 0x3f3f3f
using namespace std;
char str[100010],s[100010];
int n,fail[100010],mark[100010];
void getFail(char *s){
    int i,j;
    fail[0] = fail[1] = 0;
    for(i = 1; s[i]; i++){
        j = fail[i];
        while(j && s[i] != s[j]) j = fail[j];
        fail[i+1] = s[i] == s[j]?j+1:0;
    }
}
void kmp(char *str,char *p){
    memset(mark,-1,sizeof(mark));
    int i,j;
    for(i = j = 0; str[i]; i++){
        while(j && str[i] != p[j]){j = fail[j];}
        if(str[i] == p[j]) j++;
        if(j && mark[j] == -1) mark[j] = i;
    }
}
bool Find(char *str,char *p){
    int i,j,len = strlen(p),slen = strlen(str);
    for(i = j = 0; str[i]; i++){
        while(j && str[i] != p[j]) j = fail[j];
        if(str[i] == p[j]) j++;
        if(j && mark[len-j] != -1 && mark[len-j] < slen-i-1)
            return true;
    }
    return false;
}
int main(){
    int ans;
    while(~scanf("%s",str)){
        scanf("%d",&n);
        ans = 0;
        while(n--){
            scanf("%s",s);
            getFail(s);
            kmp(str,s);
            reverse(str,str+strlen(str));
            reverse(s,s+strlen(s));
            getFail(s);
            if(Find(str,s)) ans++;
            reverse(str,str+strlen(str));
        }
        printf("%d
",ans);
    }
    return 0;
}