Period(kmp简单的应用 next从零开始 )

最新推荐文章于 2019-09-08 09:55:57 发布

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本文介绍了一种用于检测字符串前缀是否具有周期性的算法。该算法通过构建字符串的next数组来高效判断每个前缀是否可以表示为某个字符串的多次重复，并输出所有具有周期性的前缀及其周期。

Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 16822		Accepted: 8095

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

#include<stdio.h>
#include<string.h>
#define maxn 1000100
int next[maxn];
char s[maxn];
void get_next(char *s)
{
    int len1=strlen(s);

    int i=0,j=-1;
    next[0]=-1;
    while(i<len1)
    {
        if(j==-1||s[i]==s[j])
        {
            next[++i]=++j;
        }
        else j=next[j];
    }

}

int main()
{
    int n,i,j;
    int cnt=0;
    while(~scanf("%d",&n)&&n)
    {
        scanf("%s",s);
        get_next(s);
        printf("Test case #%d\n",++cnt);
        for(i=1;i<=n;i++)
        {
           int len=i-next[i];
            if(i!=len&&i%len==0)
            {
                printf("%d %d\n",i,i/len);
            }
        
        }
        printf("\n");
    }
    return 0;
}