Period

最新推荐文章于 2021-05-20 15:57:31 发布

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最新推荐文章于 2021-05-20 15:57:31 发布

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分类专栏： KMP 文章标签： KMP

本文链接：https://blog.youkuaiyun.com/u013488894/article/details/19246379

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本文探讨了如何识别给定字符串前缀的周期性，包括输入解析、算法实现及输出格式，通过实例深入理解字符串周期性的概念。

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Period

Time Limit: 1000MS Memory limit: 65536K

题目描述

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.

输入

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.

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.

示例输入

3
aaa
12
aabaabaabaab
0

示例输出

Test case #1

2 2

3 3

Test case #2

2 2

6 2

9 3

12 4

这个题和 Power Strings其实是一个性质，只不过要把所有可能的子字符串写出来，只需在运行时只要有可能的就输出。。。

#include<stdio.h>
#include<string.h>
char t[110000000];
int next[11000000];
int main()
{
    int n,m,i,j,k;
    int len1,len2;
    int s=1;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        getchar();
        for(k=0; k<n; k++)
        {
            scanf("%c",&t[k]);
        }
        printf("Test case #%d\n",s++);
        i=0,j=-1;
        next[0]=-1;
        while(i<n)
        {
            if(j==-1||t[i]==t[j])
            {
                i++;
                j++;
                if(t[i]!=t[j]) next[i]=j;
                else next[i]=next[j];
                if(i%(i-j)==0&&i/(i-j)>1)
                {
                    int l=i/(i-j);
                    printf("%d %d\n",i,l);
                }
            }
            else j=next[j];
        }
        printf("\n");
    }
    return 0;
}