poj 1961 Period(未优化KMP的next数组)

poj 1961 Period(未优化KMP的next数组)
Time Limit: 3000ms Memory Limit: 65536kB

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 AK,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

Source
Southeastern Europe 2004

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本题复习了KMP的next[]数组求法，
使用课本里的标准程序:

void find_next(){    int i=0,j=-1;    next[i]=j;    while (i<l)    {        while ((j!=-1) && (s[i]!=s[j]))            j=next[j];        i++,j++;        next[i]=j;//可以优化    }    return;}

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Accepted    8960kB  58ms    505 B   G++

#include<stdio.h>const int LEN=1000005;int l;char s[LEN];int next[LEN];void find_next(){    int i=0,j=-1;    next[i]=j;    while (i<l)    {        while ((j!=-1) && (s[i]!=s[j]))            j=next[j];        i++,j++;        next[i]=j;    }    return;}int main(){    int cases=0;    while (scanf("%d\n",&l)&&l)    {        scanf("%s\n",s);        find_next();        cases++;        printf("Test case #%d\n",cases);        for (int i=1;i<=l;i++)            if ((next[i]!=0) && (i%(i-next[i])==0))                printf("%d %d\n",i,i/(i-next[i]));        printf("\n");    }    return 0;}