HDU 1358 Period

HDU 1358 Period
PeriodTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 3749 Accepted Submission (s): 1841


Problem Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, aggressive), we want to know whether the prefix is a periodic string. that is, for each I (2 <= I <= N) we want to know the largest K> 1 (if there is one) such that the prefix of S with length I can be written as AK, that is A concatenated K times, for some string. of course, we also want to know the period K.

Input The input file consists of several test cases. each test case consists of two lines. the first one contains N (2 <= N <= 1 000 000)-the size of the string S. the second line contains the string S. the input file ends with a line, having the number zero on it.

Output For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length I that has a period K> 1, output the prefix size I and the period K separated by a single space; the prefix sizes must be in increasing order. print a blank line after each test case.

Sample Input

3aaa12aabaabaabaab0

Sample Output

Test case #12 23 3Test case #22 26 29 312 4

Uh .. The meaning is that, starting from the second position, if there is the same loop string before, the output position is the following table at this time and the number of loop strings is output. Idea: run a KMP (only the value of the moving distance between a certain position and the current position is the 1/m of the current position, when the value in the next array is not 0, a loop string appears. (I do not know if this is true ).. ..

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          Using namespace std; # define mod 1000000007 # define inf 0x7f7f7f7fint Next [1000005]; char s [1000005]; // optimistic about the question, the array opens 100 W, I opened 10WRE many times .. Int l; void getNext () {int I, j; I = 0; j =-1; Next [I] = j; while (I
         
           > N, n) {scanf ("% s", s); l = strlen (s); memset (Next, 0, sizeof (Next); getNext (); printf ("Test case # % d \ n", ++ Case); int k; // for (k = 0; k
          
            So, find