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最新推荐文章于 2019-08-25 15:56:33 发布

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最新推荐文章于 2019-08-25 15:56:33 发布

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分类专栏： KMP

本文链接：https://blog.youkuaiyun.com/u011497404/article/details/9617939

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本文探讨了KMP算法中next[]数组的作用，特别是在寻找字符串的最小循环节方面的应用。通过实例分析，展示了如何利用特定条件下的i%(i-next[i])==0来判断是否存在最小循环节，并计算其长度。

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KMP的精髓就在于next[]数组，求最小循环节是KMP的一个活用，当i%(i-next[i])==0，最小循环节存在，为i/(i-next[i])

这道题就是求前i个字符有几个最小循环节。

Period

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 18 Accepted Submission(s) : 9

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

#include <iostream>
#include <cstring>
using namespace std;
int next[1000002];
char a[1000005];
int main()
{
    int len,n,i,j,m;
    m=1;
    while(cin>>n&&n)
    {

        cin>>a;
        len=strlen(a);
        cout<<"Test case #"<<m++<<endl;
        i=0,j=-1,next[0]=-1;
        while(i<n)
        {
            if(j==-1||a[i]==a[j])
            {
                i++,j++;
                if(i%(i-j)==0&&i/(i-j)>1)
                    cout<<i<<" "<<i/(i-j)<<endl;
                if(a[i]!=a[j])
                    next[i]=j;
                else
                    next[i]=next[j];
            }
            else
            {
                j=next[j];
            }
        }
        cout<<endl;
    }
}