poj 2752 Seek the Name, Seek the Fame ---KMP,字符串处理

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Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8194		Accepted: 3844

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

题目大意 ： 给一个字符串 输出所有 前缀和后缀相等时 的 长度。
题目思路： KMP 的next数组的一个应用。next数组记录的就是当前字符 之前的 字符串 最长的相等的前后缀 的长度。
具体可以看看kmp中next 的资料。 这题是kmp 的基础

//Memory: 4120 KB		Time: 172 MS
//Language: C++		Result: Accepted

#include
#include
const int MAX=500000;
char a[MAX]; int next[MAX];
void print(int len,int m)     // 逆序输出
{
	if(len==0) return;
	print(next[len],1);
	printf("%d",len);
	if(!m) printf("\n");
	else printf(" ");
}
int main()
{
//	freopen("in.txt","r",stdin);
	int i,j,len;
	while(scanf("%s",a)!=EOF)
	{
		len=strlen(a);
		next[0]=-1;
		i=0; j=-1;
		while(i		{
			if(j==-1 || a[i]==a[j])
			{
				i++; j++;
				next[i]=j;
			}
			else j=next[j];
		}
		print(len,0);
	}
	return 0;
}