Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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： 从字符串第二个位置开始，前面的字符是否是循环的字符串，如果是，输出当前位置及其循环的个数。

又是kmp的getnext（）函数循环节的考察

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<queue>
#include<stack>
#include<set>
#include<map>

using namespace std;

char  b[1000500];
int Next[1000500];

void get_next(char b[], int m)
{
	int i = 0,j = -1;
	memset(Next,0,sizeof(Next));
	Next[0] = -1;
	while (b[i])
	{
		if (j == -1 || b[i] == b[j])
		{
			++i;
			++j;
			Next[i] = j;
		}
		else
			j = Next[j];
	}
}

int main()
{
	int cases = 1, n, m, i, j;

	while (scanf("%d",&n)!=EOF && n)
	{
		scanf("%s",b);
		get_next(b, n);
		printf("Test case #%d\n",cases++);

		for (int i = 2; b[i-1]; i++)
		{
			int t = Next[i];
			int s = i - t;
			if (i % s == 0 && i/s>1)
			{
				printf("%d %d\n",i,i/s);
			}
		}
		printf("\n");
	}
	return 0;
}