Power Strings

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27403   Accepted: 11468 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n). Input Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case. Output For each s you should print the largest n such that s = a^n for some string a. Sample Input abcd aaaa ababab . Sample Output 1 4 3 Hint This problem has huge input, use scanf instead of cin to avoid time limit exceed. Source Waterloo local 2002.07.01 /** 题目的意思就是求字符串中最大重复子串，即a^n=s求a最大时n的值。 用kmp的next数组来求解，next数组是用来求最大前缀和后缀子串的。 则求出s的值，即s的最长前缀子串和后缀子串相等next[len]的值。 If len%(len-next[len])==0 则len/(len-next[len])就是重复子串的个数，而最大前缀的补就为a。 原因如下 1 2 3 4 5 6 7 len-next[len] 因为len%(len-next[len])==0所以可以假设将s等分为7个则7和6就相等。而1-6和2-7相等6(2-7)和5(1-6)就相等， 即5和7就相等，依次类推就都相等了…… **/ #include #include using namespace std; #define maxn 1001000 char s[maxn]; int len, next[maxn]; void get_next() { int i=0,j=-1; next[0]=-1; while(i