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poj2406--Power Strings（KMP求最小循环节）
Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 33178 | Accepted: 13792 |
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
abcdaaaaababab.
Sample Output
143
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
Waterloo local 2002.07.01
用next数组求出整个数组的最大前缀，如果整个串是用循环节组成的，那么 n - next[n] 也就是最小循环节，验证最小循环节会被n整出。
#include <cstdio>#include <cstring>#include <algorithm>int next[1100000] ;char str[1100000] ;void getnext(int l){ int j = 0 , k = -1 ; next[0] = -1 ; while(j < l) { if( k == -1 || str[j] == str[k] ) { j++ ; k++ ; next[j] = k ; } else k = next[k] ; }}int main(){ int l , m ; while(scanf("%s", str)!=EOF) { if( str[0] == '.' ) break; l = strlen(str); getnext(l) ; m = next[l]; if( l % (l-m) != 0 ) printf("1\n"); else { m = l / ( l-m ); printf("%d\n", m); } memset(str,0,sizeof(str)); } return 0;}
poj2406--Power Strings（KMP求最小循环节）
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