hdu 1358 Period

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5479    Accepted Submission(s): 2641

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

 

Recommend

JGShining

这是对KMP算法next数组的一个运用，必须要深入理解next数组；

next数组描绘的是一个字符串的对称程度（不是中心对称，即不是abcddcba这种对称，而是abcdabcd这种样子的对称）；

也就是一个字符串前缀和后缀的最长匹配；

如果当前字符串长度能够整除此长度减去当前最长匹配，那么就证明存在周期；（这里讲不太清楚，求教）

再判断一下是否只存在一个周期即可；

#include<iostream>#include<cstdlib>#include<string>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<stack>#include<queue>#include<iomanip>#include<map>#include<set>#include<functional>#define pi 3.14159265358979323846using namespace std;int N;string s;int Next[1000005];void getnext(){    int q,k;    Next[0]=0;    for(q=1,k=0;q<N;++q)    {        while(k>0&&s[q]!=s[k])            k=Next[k-1];        if(s[q]==s[k])        {            ++k;        }        Next[q]=k;    }}int main(){    int cnt=0;    while(scanf("%d",&N)!=EOF&&N)    {        ++cnt;        cin>>s;        getnext();        printf("Test case #%d\n",cnt);        for(int i=2;i<=N;++i)        {            if(i%(i-Next[i-1])==0&&i/(i-Next[i-1])>1)                printf("%d %d\n",i,i/(i-Next[i-1]));        }        printf("\n");    }    return 0;}