BNU 13174 Substring Frequency

3C. Substring Frequency

Time Limit: 1000ms

Memory Limit: 32768KB

64-bit integer IO format:  %lld      Java class name: Main

 

 

A string is a finite sequence of symbols that are chosen from an alphabet. In this problem you are given two non-empty strings Aand B, both contain lower case English characters. You have to find the number of times B occurs as a substring of A.

 

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case starts with two lines. First line contains A and second line contains B. You can assume than 1 ≤ length(A), length(B) ≤ 106.

 

Output

For each case, print the case number and the number of times B occurs as a substring of A.

 

Sample Input

4

axbyczd

abc

abcabcabcabc

abc

aabacbaabbaaz

aab

aaaaaa

aa

 

Sample Output

Case 1: 0

Case 2: 4

Case 3: 2

Case 4: 5

 

解题：裸KMP的使用。。。。。。。。。。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #define LL long long

13 #define INF 0x3f3f3f3f

14 using namespace std;

15 int fail[1000010];

16 char str[1000020],p[1000010];

17 void getFail(int &len){

18     fail[0] = fail[1] = 0;

19     len = strlen(p);

20     for(int i = 1; i < len; i++){

21         int j = fail[i];

22         while(j && p[i] != p[j]) j = fail[j];

23         fail[i+1] = p[i] == p[j]?j+1:0;

24     }

25 }

26 int main(){

27     int t,i,j,len,ans,k = 1;

28     scanf("%d",&t);

29     while(t--){

30         scanf("%s%s",str,p);

31         getFail(len);

32         j = ans = 0;

33         for(i = 0; str[i]; i++){

34             while(j && str[i] != p[j]) j = fail[j];

35             if(str[i] == p[j]) j++;

36             if(j == len) ans++;

37         }

38         printf("Case %d: %d\n",k++,ans);

39     }

40     return 0;

41 }

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