【字符串入门专题1】F

Seek the Name, Seek the Fame

Description

The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm: 

Step1. Connect the father's name and the mother's name, to a new string S. 
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S). 

Example: Father='ala', Mother='la', we have S = 'ala'+'la' = 'alala'. Potential prefix-suffix strings of S are {'a', 'ala', 'alala'}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:) 

Input

The input contains a number of test cases. Each test case occupies a single line that contains the string S described above. 

Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000. 

Output

For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby's name.

Sample Input

ababcababababcabab
aaaaa

Sample Output

2 4 9 18
1 2 3 4 5

题意：按从小到大的顺序输出字符串前缀和后缀相等的长度.

思路：利用next数组的性质，从后往前滚动查找，因为s[i]必定等于s[next[i]],最后输出.

这道题真是超时超到爆炸，去掉判断语句，沿用小伙伴的方法，自己就a了，真尴尬~~~

超级喜欢递归调用输出，感觉就像黑科技一样（好像有点夸张了），不用新的数组存值，避开了我数组范围总是定小的毛病

#include<stdio.h>
#include<string.h>
#define N 510000
char s[N];
int NEXT[N],num[N];

void get_next()
{
	int j,k;
	NEXT[0] = k = -1;
	j = 0;
	while(s[j]!='')
	{
		if(k==-1||s[k]==s[j])
			NEXT[++j] = ++k;
		else
			k = NEXT[k];
	}
	
	return;
}

void print(int n)
{
	if(NEXT[n] > 0)
	{
		print(NEXT[n]);
		printf("%d ",NEXT[n]);
	}
}
int main()
{
	int i,j;
	while(scanf("%s",s)!=EOF)
	{
		get_next();
		print(strlen(s));
		printf("%d
",strlen(s));
	}
	return 0;
}