HDU1358 Period

题目链接：HDU1358

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5769    Accepted Submission(s): 2786

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 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 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

   
   
   
   

    
    
    
    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：给一个字符串，问从串首到那个位置截取的字串是以周期循环的，循环了多少次。

用kmp 求周期的方式，求完next数组后再从头遍历一遍。

//
//  main.cpp
//  HDU1358（2）
//
//  Created by teddywang on 16/4/23.
//  Copyright © 2016年 teddywang. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[1000010];
int nexts[1000010];
int len;

void getnext()
{
    int i=0,j=-1;
    nexts[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])
        {
            nexts[++i]=++j;
        }
        else j=nexts[j];
    }
}

int main()
{
    int T,l=1;
    while(cin>>T&&T)
    {
        scanf("%s",s);
        len=T;
        getnext();
        printf("Test case #%d\n",l++);
        for(int i=2;i<=len;i++)
        {
            int m=i-nexts[i];
            if(i%m==0&&m!=i)
            {
                cout<<i<<" "<<i/m<<endl;
            }
        }
        cout<<endl;
    }
}