1358 Period

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5990    Accepted Submission(s): 2917

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

题目大意：给你一个字符串s，求s的那些前缀可以表示成A^K（k>1）次方。

比如样例二，s = aabaabaabaab，那么s 的前缀有

a                                     可以表示成 a^1，但是因为k 要大于1，所有不符合条件

aa                                  a^2

aab                               aab^1

......

aabaabaabaab         aab^4

题目的意思就是这样啦

思路：这个可以理解为求字符串的每个前缀的循环节，然后用长度除以循环节的长度就是答案了

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<map>#define LL long longusing namespace std;const int maxn = 1000000+5;int Next[maxn];void getNext(string s){    int ls=s.size();    Next[0]=-1;    Next[1]=0;    for(int i=2;i<=ls;i++){        if(s[i-1]==s[Next[i-1]]) Next[i]=Next[i-1]+1;        else{            int t=Next[i-1];            while(s[i-1]!=s[t]){                t=Next[t];                if(t==-1) break;            }            Next[i]=t+1;        }    }}int main(){    int n,ls,c=0;    string s;    while(cin>>n){        if(n==0) break;        cin>>s;        getNext(s);        ls=s.size();        cout<<"Test case #"<<++c<<endl;        for(int i=0;i<=ls;i++){            if(Next[i]>0){                if(i%(i-Next[i])==0){                    cout<<i<<' '<<i/(i-Next[i])<<endl;                }            }        }        cout<<endl;    }    return 0;}