Period

最新推荐文章于 2021-11-23 13:40:30 发布

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最新推荐文章于 2021-11-23 13:40:30 发布

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本文链接：https://blog.youkuaiyun.com/weixin_43841060/article/details/85081782

本文探讨了KMP算法在处理周期字符串问题上的应用，通过分析前缀与字符串的重复模式，介绍了如何利用KMP算法的next数组来找出字符串的所有可能周期。并通过一道具体题目，展示了算法的具体实现过程。

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p.s. KMP.next数组循环节应用

Period

Time Limit: 1000 ms Memory Limit: 65536 KiB

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

题目链接：

http://acm.sdut.edu.cn/onlinejudge2/index.php/Home/Contest/contestproblem/cid/2710/pid/2476

#include <bits/stdc++.h>

using namespace std;

int nextnum[1000010];
char s[1000010];
int n,num=0;

void get_next(char s[]){
    int i=0,j=-1,len=strlen(s);
    nextnum[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])
        {
            i++;
            j++;
            nextnum[i]=j;
        }
        else
        j=nextnum[j];
    }
}

int main()
{
    while(cin >> n&&n)
    {
        cin >> s;
        num++;
        get_next(s);
        cout << "Test case #" << num << endl;
        for(int i=2;i<=n;i++)
        {
            if(i%(i-nextnum[i])==0&&i/(i-nextnum[i])>=2)
            cout << i << ' ' << i/(i-nextnum[i]) << endl;
        }
        cout << endl;
    }
    return 0;
}