hdu 1358 Period(求循环节)_d - period hdu

本文链接：https://blog.csdn.net/acm_fighting/article/details/51072662

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5684 Accepted Submission(s): 2738

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:给你一个字符串S,问你有多少个i使得S的前i个字符组成的前缀是某个字符串重复K次得到,输出所有存在K的i和对应的K。

比如字符串aabaabaabaab,只有当i=2,6,9,12时K存在,且分别为2,2,3,4.

思路:当i%(i-Next[i])==0&&Next[i]!=0前i个字符可以组成一个周期串。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1100000;
int n,m;
char pattern[maxn];
int Next[maxn];

void get_next(){
    int i=0,j=-1;
    Next[0]=-1;
    while(i<m){
        if(j==-1||pattern[j]==pattern[i]){
            j++,i++;
            Next[i]=j;
            if(i%(i-Next[i])==0&&Next[i]!=0)
                printf("%d %d\n",i,i/(i-Next[i]));
        }
        else
            j=Next[j];
    }
}

int main(){
    int case1=1;
    while(scanf("%d",&m)!=EOF){
        if(m==0)
            break;
        scanf("%s",pattern);
        printf("Test case #%d\n",case1++);
        get_next();
        printf("\n");
    }
    return 0;
}