利用kmp裏的next數組的一道題

題目

E.T. Inc. employs Maryanna as alien signal researcher. To identify possible alien signals and background noise, she develops a method to evaluate the signals she has already received. The signal sent by E.T is more likely regularly alternative.

Received signals can be presented by a string of small latin letters a to z whose length is $N$. For each $X$between $1$ and $N$ inclusive, she wants you to find out the maximum length of the substring which can be written as a concatenation of $X$ same strings. For clarification, a substring is a consecutive part of the original string.

Input

The first line contains $T(T\le 200)$, the number of test cases. Most of the test cases are relatively small. $T$ lines follow, each contains a string of only small latin letters a - z, whose length $N$ is less than $1000$, without any leading or trailing white spaces.

Output

For each test case, output a single line, which should begin with the case number counting from $1$, followed by $N$integers. The $X$-th ($1$-based) of them should be the maximum length of the substring which can be written as a concatenation of $X$ same strings. If that substring doesn't exist, output $0$ instead. See the sample for more format details.

Hint

For the second sample, the longest substring which can be written as a concatenation of $2$ same strings is noonnoon, oonnoonn, onnoonno, nnoonnoo, any of those has length $8$; the longest substring which can be written as a concatenation of $3$ same strings is the string itself. As a result, the second integer in the answer is $8$ and the third integer in the answer is $12$.

樣例輸入

2
arisetocrat
noonnoonnoon

樣例輸出

Case #1: 11 0 0 0 0 0 0 0 0 0 0
Case #2: 12 8 12 0 0 0 0 0 0 0 0 0

題目大意

  輸入t個字符串，輸出從1到strlen(str)的循環節的最大長度

 例如第二個樣例，循環節爲1個的可以把自己看成循環節最大長度就是12

  循環節爲2個的情況是noonnoon , oonnoonn , onnoonno, nnoonnoo,長度是8

 循環節是3個的情況只有noonnoonnoon,這時候的單個循環節就是noon

 

解題思路

 利用next數組，next數組處理之後的跳變位置，我們利用next數組的特性，next[k] 與 next[next[k]]字符串相同的特性就可以找出最小的循環節，然後小的循環節如果能疊加變成一個大的循環節，例如ababababab,ab是一個小的循環節，abab則是疊加之後形成的一個大的循環節，同樣利用next數組的特性處理即可，通過這題加深了對next數組的理解

代碼

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std ;
const int maxn = 1000 + 10 ;
int nex[maxn] , c[maxn] ;
int case_ = 0 ;
void get_nex(char* s , int len)
{
    int  j = 0 ;
    nex[0] = 0 ;
    nex[1] = 0 ;
    for(int i = 2 ; i <= len ; i++)
    {
        while(j && s[j+1] != s[i])
            j = nex[j] ;
        if(s[j+1] == s[i])
            j++ ;
        nex[i] = j ;
    }
}
int main()
{
    int t ;
    char str[maxn] ;
    scanf("%d" , &t) ;
    while(t--)
    {
        memset(c , 0 , sizeof(c)) ;
        scanf("%s" , str+1) ;
        int len = strlen(str + 1) ;
        for(int i = 0 ; i < len ; i++)
        {
            int m = len - i ;
            get_nex(str + i , m) ;
            for(int j = 1 ; j <= m ; j++)
            {
                int tmp = j ;
                while(tmp) ///這裏處理 上文提及的疊加處理
                {
                    tmp = nex[tmp] ;
                    if( j % (j - tmp) == 0)
                    {
                        int tt = j / (j - tmp) ;
                        c[tt] = max(c[tt] , j) ;
                    }
                }
            }
        }
        printf("Case #%d:", ++case_);
        for (int i = 1 ; i <= len ; i++)
            printf(" %d", c[i]);
        printf("\n");
    }
    return 0 ;
}