hdu1358—Period（循环串）

题目链接：传送门

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9822    Accepted Submission(s): 4693

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

3aaa12aabaabaabaab0

 

Sample Output

Test case #12 23 3Test case #22 26 29 312 4

解题思路：利用next数组，对于前k个字符，令L = k-next[k]，若k%L == 0，则L就是循环周期最长的串的长度，在图上画一下能证明。

#include <iostream>#include <algorithm>#include <stdio.h>#include <string.h>using namespace std;//hdu1358//对于前k个字符，令L = k-next[k]，若k%L == 0，则L就是循环周期最长的串的长度//在图上画一下能证明const int N = 1000100;int nt[N],n;char st[N];void GetNext(char* p){    int pLen = n+1;    nt[0] = -1;    int k = -1,j = 0;    while (j < pLen - 1){        //p[k]表示前缀，p[j]表示后缀        if (k == -1 || p[j] == p[k]){            ++k;            ++j;            nt[j] = k;        }else{            k = nt[k];        }    }}int main(){    int cas = 0;    while(~scanf("%d",&n)&&n){        scanf(" %s",st);        st[n] = '#';        GetNext(st);        printf("Test case #%d\n",++cas);        for( int i = 2 ; i <= n ; ++i ){            int len = i-nt[i];            if(nt[i] != 0 && i%len == 0){                printf("%d %d\n",i,i/len);            }        }        printf("\n");    }    return 0;}