HDU：1358-Period

Period

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)

Problem 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 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

---

解题心得：

1. 题意就是叫你找给出的字符串的每一个前缀中存在的循环节，其实很简单，就是在建立next数组的时候一边匹配，一边寻找循环节。裸找循环节。

---

#include<stdio.h>
#include<cstring>
using namespace std;
const int maxn = 1e6+100;
char s[maxn];
int next[maxn],n,t;

void cal_next()
{
    memset(next,-1,sizeof(next));
    next[0] = -1;
    int k = -1;
    for(int i=1;i<n;i++)
    {
        while(k > -1 && s[i] != s[k+1])
            k = next[k];
        if(s[i] == s[k+1])
            k++;
        next[i] = k;
        if(k == -1)//当是-1的时候肯定没有循环节
            continue;
        int len = i+1;
        if(len%(i-k) == 0)
            printf("%d %d
",len,len/(i-k));
    }
    printf("
");
}

int main()
{
    t = 1;
    while(scanf("%d",&n) && n)
    {
        scanf("%s",s);
        printf("Test case #%d
",t++);
        cal_next();
    }
}