FINDSR - Find String Roots
In mathematics, the N-th root of a number M, is a number K such that KN = M , i.e. KKK . . . K = M where K is multiplied N times.
We can translate this into strings. In string notation, the yuxtaposition is concatenation instead of multiplication. So, the N-th root of a string S is another string T such that TN = S, where T N = T T T . . . T is the string T concatenated N times. For instance, if S = “abcabcabcabc”, for N = 2 the string T = “abcabc” is the N-th root of S, while for N = 4 its N-th root is T = “abc”. Note that for N = 1 any string S is the N-th root
of S itself.
Given a string S you have to find the maximum N such that the N-th root of S exists. In the above example the answer would be 4, because there is no N-th root of S = “abcabcabcabc” for N > 4.
The input contains several test cases, each one described in a single line. The line contains
a non-empty string S of at most 105 characters, entirely formed of digits and lowercase
letters. The last line of the input contains a single asterisk (“*”) and should not be
processed as a test case.
For each test case output a single line with the greatest integer N such that there exists
a string T that concatenated N times is equal to S.
use Z algo ....
spojtoolkit giving wrong answers for few sample inputcase
Similar problem - https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1239 ... Enjoy Coding
Pleased to find out that O(n) passes comfortably in Python. Comments were helpful - learn a new bit about KMP instead of wasting time on breaking open doors.
vijay kumar paliwal:
Same as PERIOD !! Indeed PERIOD is generalization of this..
afetr 3 WA . finally AC .
-_-Last edit: 2017-01-16 19:20:06
is the complexity N logN??
Almost the same as spoj.com/problems/PERIOD xD xD ; easy KMP