Period（NEXT）

字符串周期性检测

最新推荐文章于 2020-12-16 11:50:00 发布

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Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11087 Accepted Submission(s): 5207

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 AK , 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

思路：
这个题就是使用next[]十足，next[i]是从0~i的字符串的前缀和后缀相同的长度。
i+1-next[i]就是周期长度，如果(i+1)%temp==0 && (i+1)/temp>1，满足，输出。

#include <iostream>
using namespace std;
#define MAXN 1000005
int Next[MAXN];
string s;

void makeNext(string s1,int Next[]){
    int q,k;
    int m=s1.length();
    Next[0]=0;
    for(int q=1,k=0;q<m;q++){   
        while(k>0 && s1[q]!=s1[k])
            k=Next[k-1];
        if(s1[q]==s1[k]){
            k++;
        }
        Next[q]=k;
    }
}

int main(){
    std::ios::sync_with_stdio(false);  
    int n,t=0;
    while(cin>>n && n!=0){
        t++;
        cin>>s;
        makeNext(s,Next);
        cout<<"Test case #"<<t<<endl;
        for(int i=1;i<n;i++){
            int temp=i+1-Next[i];
            if((i+1)%temp==0 && (i+1)/temp>1)
                cout<<i+1<<" "<<(i+1)/temp<<endl;
        }
        cout<<endl;
    }
    return 0;

}