poj 1961Period

Period

Time Limit: 3000MS

Memory Limit: 30000K

Total Submissions: 9885

Accepted: 4511

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

[题目大意]：给定一个字符串，求到哪一位时的字串是前几位循环组成的，并求出循环次数。

 

[分析]：利用KMP来求解。在KMP中有一个p数组记录当前字符的它的上一个位置，且保证这一段一定是来连续的。所以如果i处有循环则i到p[i]的长度即为循环节长度，所以如果i 
mod  (i-p[i])=0则说明有循环，i div (i-p[i])就是循环次数。

[代码]：
#include<iostream>
#include<string>
using namespace std;
char s[1000005];
int next[1000005];
char c[10];
int main()
{
    int t,i,j,k,n,len;
    t=0;
    while(scanf("%d",&n)!=EOF)
    {
       gets(c);
       gets(s);
       len=strlen(s);
       next[0]=-1;
       j=0;k=-1;
       while(j<len)
       {
          if(k==-1||s[j]==s[k])
          {
            j++;
            k++;
            next[j]=k;
          }
          else k=next[k];
       }
       printf("Test case #%d\n",++t);
       for(i=2;i<=len;i++)
       {
         k=i-next[i];
         if(i%k==0&&i/k>1) printf("%d %d\n",i,i/k);
       }
       printf("\n");
    }
    return 0;
}