Period next数组的应用

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本文介绍了一种用于检测字符串前缀是否为周期串的算法，通过构造next数组来判断每个前缀是否存在周期性，并计算出最小周期的长度及数量。文章详细解析了next数组的生成过程及其在周期检测中的应用。

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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.

输入：

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.

输出:

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.

样例输入：

3
aaa
12
aabaabaabaab
0

样例输出：

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

这题就是给出一串字符，求前缀是周期串的长度，并且求出这个周期串有几个最小周期。例如abaaba是一个周期串，最小周期串是aba，一共有两个最小周期。

这道题主要是关于next数组的应用，以第二组案例为例。

i 0 1 2 3 4 5 6 7 8 9 10 11 12

a[i] a a b a a b a a b a a b

net[i] -1 0 1 0 1 2 3 4 5 6 7 8 9

可以看出最小周期的长度时i-next[i]，所以首先要判断i%(i-next[i])==0，并且题目中说这个长度K是大于1的，所以要满足i/(i-next[j])>1 (代码中的next数组用net表示，因为用next会显示标识符不明确)

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=1000005;
char s[maxn];
int net[maxn];
int n;
void getnet()
{
    net[0]=-1;
    int j=0;
    int k=-1;
    while(j<n)
    {
        if(k==-1||s[j]==s[k])
        {
            ++j;
            ++k;
            net[j]=k;
        }
        else
            k=net[k];
    }
}
int main()
{
    int cas=1;
    while(scanf("%d",&n)!=EOF &&n)
    {
        scanf("%s",s);
        getnet();
        printf("Test case #%d\n",cas++);//不要忘记这一句
        for(int i=1;i<=n;i++)
        {
            int j=i-net[i];//最小周期的长度

            if(i%j==0&&i/j>1)//如果满足条件，就输出周期的最长度和最小周期的个数
            {
                printf("%d %d\n",i,i/j);
            }
        }
        printf("\n");
    }
    return 0;
}