HDU 1358 Period

题目：

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
题意描述：
输入串的长度N(2 <= N <= 1 000 000)及长度为N的串
计算并输出所有循环串的长度i以及该循环串中存在循环节的个数K（K>1）（或者循环周期）
解题思路：
对KMP中next[]数组的应用，重点在理解next[]数组的含义，即存储了从开头到斜上位置的串的前后缀的相似度。另外使用当前串的长度减去next[]末位得到的是该串最小循环节的长度。
那么枚举从第三个字符开始到结尾的不同的串，判断当前串是否含有一个以上的循环节并且含有整数个循环节，是的话输出串的长度和循环节的个数即可。
代码实现：

#include<stdio.h>
char s[1000010];
int next[1000010];
int n;
void get_next(char t[]);
int main()
{
    int i,c,t=1;
    while(scanf("%d",&n),n != 0)
    {
        scanf("%s",s);
        get_next(s);
        printf("Test case #%d\n",t++);
        for(i=2;i<=n;i++)
        {
            c=i - next[i];
            if(i/c > 1 && i%c==0 )
            printf("%d %d\n",i,i/c);
        } 
        printf("\n");
    }
    return 0;
}
void get_next(char t[])
{
    int i,j;
    i=0;j=-1;
    next[0]=-1;
    while(i < n)
    {
        if(j==-1 || t[i]==t[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else//少个else 
            j=next[j];
    }
    /*for(i=0;i<=n;i++)
        printf("%d ",next[i]);
    printf("\n");*/
}