hdu oj Period (kmp的应用）

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2866    Accepted Submission(s): 1433

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

                                                     

这道题開始看了好久都没懂题意，后来对着例子看了一下，把题意大体弄懂了，就是给你一个字符串。求里面的循环节的个数。从第二个字符開始，推断是否是循环串，并求出循环的周期（出现的次数）。然后我自己想開始想到字符串匹配；还认为要用二个字符串，感觉kmp还是没理解到位；又把kmp看了一下，看大神的思路，事实上这就是对next数组的一种应用；next数组事实上保存的就是字符串的相似度；保存的是其前一个字符的前缀和后缀相等的最大值，假如next[i]的值为6，那么它的前一个字符前缀和后缀最多就有6个相等。

字符的编号从0開始。这里面就能够推出一个规律；那么if(i%(i-next[i])==0)，则i前面的串为一个轮回串，当中轮回子串出现i/(i-next[i])次。

以下是ac的代码。

#include <cstdio>
#include <cstring>
int n;
int next[1000001];
char s[1000001];
void get_next()//求出next数组
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<=n)
    {
        if(j==-1 || s[i]==s[j])
        {
            ++i;
            ++j;
            next[i]=j;
        }
        else
        j=next[j];
    }
}
int main()
{
    int t,j=1;
    while(scanf("%d",&n)&&n)
    {
       scanf("%s",s);
       get_next();
       printf("Test case #%d\n",j++);
       for(int i=2;i<=n;i++)
       {
           t=i-next[i]; //关键点在这里
           if(i%t==0&&i/t>1)
           {
               printf("%d %d\n",i,i/t);
           }
       }
       printf("\n");
    }
    return 0;
}