POJ-1961-Period

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

KMP的前缀函数处理出来的前缀数组表示当当前字符失配后，要向前调到哪一个位置可以继续匹配。也就是代表着跳到的那个位置之前的所有字符与当前失配字母前的相同数量个字母是相匹配的。

也是KMP中循环节的模板题

#include <stdio.h>
#include <string.h>
#include <algorithm>
const int N=1000100;
char a[N],b[N];
int len,n[N];
void next()
{
    int i=0;
    int j=-1;
    n[0]=-1;
    while(i<len)
    {
        if(j==-1||a[i]==a[j])
        {
            i++;
            j++;
            n[i]=j;
        }
        else
            j=n[j];
    }
}
int main()
{
    int t=1;
    while(scanf("%d",&len)!=EOF&&len)
    {
        memset(n,0,sizeof(n));
        scanf("%s",a);
        next();
        printf("Text case #%d
",t++);
        for(int i=2; i<=len; i++)//n[i]代表到的失配后跳那的个位置，它之前的所有字符与
            //当前失配字母前的相同数量个字母是相匹配的，即只有这样才能成为循环节
        {//i-n[i]等于他们匹配字母数量即循环节的长度
            if(i%(i-n[i])==0&&n[i])//如果不能被整除，就说明i是由循环节和循环节的一部分结组成
            {//如果可以被整除，即说明整条字符串都是由小的循环节组成
                printf("%d %d
",i,i/(i-n[i]));
            }
        }
        printf("
");
    }
    return 0;
}