poj--1961--Period

最新推荐文章于 2019-08-25 15:56:33 发布

原创最新推荐文章于 2019-08-25 15:56:33 发布 · 303 阅读

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本文介绍了一种利用KMP算法高效求解字符串前缀周期的方法。通过解析next数组，文章详细阐述了如何找出字符串中所有前缀的循环节长度，并提供了一段简洁的C++代码实现。

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Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 18603		Accepted: 9044

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

题意：

求字符串的每个前缀的大于等于2的循环节长度

思路：

考察KMP算法中对next数组的理解，如果对KMP算法比较了解的话，很容易就能写出来

代码：

 C++ Code 

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#include<cstdio>
using namespace std;

const int MAXN = 1000010;
char str[MAXN];
int next[MAXN];
int n;
void getNext()
{
    int i, j;
    i = 0;
    j = -1;
    next[0] = -1;
    while(str[i] != '\0')
    {
        if(j == -1 || str[i] == str[j])
        {
            i++;
            j++;
            if(i % (i - j) == 0 && i / (i - j) > 1)
                printf("%d %d\n", i, i / (i - j));
            next[i] = j;
        }
        else j = next[j];
    }
}
int main()
{
    int iCase = 0;
    while(scanf("%d", &n), n)
    {
        iCase++;
        scanf("%s", &str);
        printf("Test case #%d\n", iCase);
        getNext();
        printf("\n");
    }
    return 0;
}