首页 > 代码库 > poj1961--Power Strings(kmp:求循环串的次数)
poj1961--Power Strings(kmp:求循环串的次数)
Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 33163 | Accepted: 13784 |
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 2002.07.0
#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( m == -1 ) printf("1\n"); else if(m == l) printf("%d\n", l); else if( l % (l-m) != 0 ) printf("1\n"); else { m = l / ( l-m ); printf("%d\n", m); } memset(str,0,sizeof(str)); } return 0;}poj1961--Power Strings(kmp:求循环串的次数)
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。