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。。。。。。。。。。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #define LL long long
 #define INF 0x3f3f3f3f
 using namespace std;
 int fail[];
 char str[],p[];
 void getFail(int &len){
     fail[] = fail[] = ;
     len = strlen(p);
     for(int i = ; i < len; i++){
         int j = fail[i];
         while(j && p[i] != p[j]) j = fail[j];
         fail[i+] = p[i] == p[j]?j+:;
     }
 }
 int main(){
     int t,i,j,len,ans,k = ;
     scanf("%d",&t);
     while(t--){
         scanf("%s%s",str,p);
         getFail(len);
         j = ans = ;
         for(i = ; str[i]; i++){
             while(j && str[i] != p[j]) j = fail[j];
             if(str[i] == p[j]) j++;
             if(j == len) ans++;
         }
         printf("Case %d: %d\n",k++,ans);
     }
     return ;
 }