CUGB 전문 훈련의 데이터 구조: D - Period (kmp 의 next 판단 순환 절)

D - Period
Time Limit:3000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u
Submit 
Status
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

판단 순환: i% (i - next (i) = 0 은 순환 절 이 있다 는 것 을 설명 하고 순환 수 는 i / (i - next (i) 이다.

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d
",a)
#define f(i,a,n) for(i=a;i<n;i++)
#define F(i,a,n) for(i=a;i<=n;i++)
#define MM 1000005
#define MN 1005
#define INF 10000007
using namespace std;
char str1[MM];
int next[MM],len1;
void getnext()
{
    int k=-1,j=0;
    next[0]=-1;
    while(j<len1)
    {
        if(k==-1||str1[j]==str1[k]) next[++j]=++k;
        else k=next[k];
    }
}
int main()
{
    int n,k=1;
    while(scanf("%d",&n)&&n)
    {
        scanf("%s",str1);
        len1=strlen(str1);
        getnext();
        printf("Test case #%d
",k++);
        for(int i=1;i<=n;i++)
        {
            int l=i-next[i];
            if(i!=l&&i%l==0) printf("%d %d
",i,i/l);
        }
        puts("");
    }
    return 0;
}