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POJ 1961 period

2024-08-01 04:28:57   218人阅读

Period
Time Limit: 3000MS Memory Limit: 30000K
Total Submissions: 13584 Accepted: 6396

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 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：求出不同的循环节下循环的长度和循环节个数

根据next数组的性质即 i-next[i] 为一个循环节的长度


#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#define N 1000009

using namespace std;

int n;
char a[N];
int next[N];
int len;

void getnext()
{
    int i,j;
    i=0;
    j=-1;
    next[0]=-1;

    while(i<n)
    {
        if(j==-1||a[i]==a[j])
           {
               i++;
               j++;
               next[i]=j;
           }
           else
           j=next[j];
    }
}

int main()
{
    int ca=1;

    while(scanf("%d",&n),n)
    {
        scanf("%s",&a);
        getnext();

//      for(int i=0;i<=n;i++)
//        cout<<next[i]<<" ";



         cout<<"Test case #"<<ca++<<endl;

         for(int i=1;i<=n;i++)//判断多种循环节的情况
         {
             len=i-next[i];
             if(len!=i&&i%len==0)//遇到循环就输出
             cout<<i<<" "<<i/len<<endl;
         }
         cout<<endl;
         
    }
    return 0;
}





POJ 1961 period


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