poj1961 Period(KMP)

C - Period

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Time Limit:3000MS    
Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

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Status Practice POJ 1961

Appoint description:
System Crawler (2016-05-10)

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

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

第二次做KMP算法了  可惜的是我见到了都不知道是要用KMP算法做  唉

上次做也没有深刻理解next数组的意义  做了这道题 让我理解到 （仅仅是我的理解，错了别喷我。。）next数组的作用就是在匹配主串的时候

遇到了不相同的字符 通过next数组快速跳过已经匹配过的字符。具体解释请看这位大牛写的

做了这道题也知道了next数组可以找循环节 唉 真的想不到

#include <stdio.h>
int next[1000000+10];
char str[1000000+10];
int n;
void set_next()
{
	int i=-1;
	int j=0;
	next[0]=-1;
	while(j<n)
	{
		if(i==-1||str[i]==str[j])
		{
			i++;j++;
			next[j]=i;
		}
		else
		i=next[i];
	}
}
int main()
{
	int t=0;
	while(~scanf("%d",&n)&&n)
	{
		scanf("%s",str);
		set_next();
		printf("Test case #%d\n",++t);
		for(int i=1;i<=n;i++)
		{
		//	printf("%d\n",next[i]);
			int length=i-next[i];
			if(i!=length&&i%length==0)
			printf("%d %d\n",i,i/length);
		}
		printf("\n");
	}
}