【字符串入门专题1】D

Period

Problem 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 file 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

3
aaa
12
aabaabaabaab
0

 

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

 

题意：按从小到大的顺序每行输出字符串的循环节长度和最小循环节的出现次数

思路：嘻嘻嘻，终于用上最小循环节的总结啦，我是最小循环节总结哦~~，这道题虽然没有用上我喜欢的递归调用输出函数，但是呐，一扫之前那道题带给我糟糕的心情，我就是那道让cdn超时到抓狂的题，总的来说还是比较简单的。

#include<stdio.h>
#include<string.h>
#define N 1100000
char s[N];
int next[N];
void get_next()
{
	int k,j;
	next[0] = k = -1;
	j = 0;
	while(s[j]!='')
	{
		if(k == -1||s[k]==s[j])
			next[++j] = ++k;
		else
			k = next[k];
	}
	return;
}
int main()
{
	int n,i;
	int t = 0;
	while(scanf("%d",&n),n!=0)
	{
		t++;
		getchar();
		scanf("%s",s);
		printf("Test case #%d
",t);
		get_next();
		for(i = 0; i <= n; i ++)
		{//判断循环节的条件
			if(i%(i-next[i]) == 0&&(next[i] != 0&&next[i]!=-1))
				printf("%d %d
",i,i/(i-next[i]));
		}
		printf("
");
	}
	return 0;
}