3C. Substring Frequency
Time Limit: 1000ms
Memory Limit: 32768KB
64-bit integer IO format: %lld Java class name: MainA 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 ",k++,ans); 39 } 40 return 0; 41 }