Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2120 Accepted Submission(s): 1039

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

aaa

aabaabaabaab

Sample Output

Test case #1

2 2

3 3

Test case #2

2 2

6 2

9 3

12 4

Recommend JGShining

#include<stdio.h>

#include<string.h>

#define gs 1000010

int next[gs];

char str[gs];

void GetNext(char * str,int * next)

{

    int N=strlen(str+);

    next[]=;

    int j=;

    for(int i=;i<=N;i++)

    {

        while(j> && str[j+]!=str[i]) j=next[j];

        if(str[j+]==str[i]) j+=;

        next[i]=j;

    }

}

int main()

{

    int N,cas=;

    while (scanf("%d",&N)!=EOF)

    {

        if (N==) return ;

        cas++;

        printf("Test case #%d\n",cas);

        scanf("%s",str+);

        GetNext(str,next);

        for (int i=;i<=N;i++)

        if (i%(i-next[i])== && i/(i-next[i])>) printf("%d %d\n",i,i/(i-next[i]));

        printf("\n");

    }

    return ;

}

巴特西

Period[HDU1358]

Period

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