POJ 1961：Period

最新推荐文章于 2021-07-19 17:07:24 发布

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最新推荐文章于 2021-07-19 17:07:24 发布

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原文链接：http://www.cnblogs.com/lightspeedsmallson/p/4785873.html

本文介绍一种算法，用于判断给定字符串的所有前缀是否具有周期性特征，并找出最大周期K。通过输入不同长度的字符串，算法能输出所有符合条件的前缀长度及其周期。

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Period
Time Limit: 3000MS Memory Limit: 30000K
Total Submissions: 14280 Accepted: 6773

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 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

POJ2406与这道题一个意思，就是这道题细化了一点。

代码：

#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;

char a[1000005];
int next1[1000005];

void cal()
{
    int len = strlen(a);
    int i,j=-1;
    next1[0]=-1;
    for(i=0;i<len;)
    {
        if(j==-1||a[i]==a[j])
        {
            i++;
            j++;

            next1[i]=j;
        }
        else
        {
            j=next1[j];
        }
    }
}

int main()
{
    int len,count=1;
    while(cin>>len)
    {
        if(len==0)
            break;

        cin>>a;
        cal();

        int i;
        cout<<"Test case #"<<count++<<endl;

        for(i=2;i<=len;i++)
        {
            if(i%(i-next1[i])==0 && i/(i-next1[i])>=2)
                cout<<i<<" "<<i/(i-next1[i])<<endl;
        }

        cout<<endl;
    }
    return 0;
}