POJ 1961 Period

2019-11-06 06:33:53

字體：大中小

供稿：網(wǎng)友

Description

For each PRefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), 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 A^K ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input 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

3aaa12aabaabaabaab0Sample Output
Test case #12 23 3Test case #22 26 29 312 4Source
Southeastern Europe 2004
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
KMP~
求出next[i]，然后枚舉截止位置i，則循環(huán)節(jié)的長(zhǎng)度now=i-next[i]，如果now!=i，則表明循環(huán)了不止一次，如果同時(shí)i%now==0（即完整循環(huán)），直接輸出即可~
#include<cstdio>#include<cstring>int len,next[1000001],k,totcase,now;char s[1000001];int main(){	while(scanf("%d",&len)==1 && len)	{		scanf("%s",s+1);k=0;		printf("Test case #%d/n",++totcase);		for(int i=2;i<=len;i++)		{			while(k && s[i]!=s[k+1]) k=next[k];			if(s[i]==s[k+1]) k++;next[i]=k;		}		for(int i=1;i<=len;i++)		{			now=i-next[i];			if(now!=i && !(i%now)) printf("%d %d/n",i,i/now);		}		printf("/n");	}	return 0;}