pku2406kmp

来源：互联网发布：韩国美臀小姐走红网络编辑：程序博客网时间：2024/06/12 00:36

题目来源：http://poj.org/problem?id=2406

题目分类：KMP next数组

此题心得：加深了对next数组的理解

作者：王耀宣

时间：2011-7-21

Power Strings

Time Limit: 3000MS

Memory Limit: 65536K

Total Submissions: 17868

Accepted: 7460

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd

aaaa

ababab

.

Sample Output

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

[Submit] [Go Back] [Status] [Discuss]

Home Page Go Back To top

题意与分析

题目：给你一个串，让你在这个串当中找到一个最长的前缀串，保证这个串是其某一个前缀s的k倍，也就是由几个s连接起来构成，求这个k的值最大能多大。

分析：其实就是利用了KMP的next数组的性质。对每个位置，next数组都指向前一个可能匹配的位置。对i处，next[i]指向的位置作为结尾的就能匹配以i为结尾的前缀的有用部分。如果一个串能被划分成几个字串的连接，则next[len]一定是指向倒数第二个的位置，那么只有len%(len-next[len]为0时才能保证是由一个串不相交连接而成的。。。

源代码

·                //pku 2046

·                #include<iostream>

·                using namespace std;

·

·                const int N=5000100;

·                char s[N];

·                int a[N], next[N], len, n, m, flag;

·

·                void getnext(char *s)

·                {

·                      int j, k;

·                      j = 0;

·                      k = -1;

·                      next[0] = -1;

·                      while(j<len)

·                      {

·                            if(k==-1 || s[j]==s[k])

·                            {

·                                  j++;

·                                  k++;

·                                  next[j] = k;

·                            }

·                            else

·                                  k = next[k];

·                      }

·                }

·

·                int main()

·                {

·                      int i, j, k, cas, cas1=1;

·                      int ans;

·                      while(scanf("%s", s)!=EOF)

·                      {

·                            if(s[0]=='.')

·                                  break;

·                            ans = 0;

·                            len = strlen(s);

·                            getnext(s);

·                            //for(i=0; i<=len; i++)

·                            //    printf("%d ", next[i]);

·                            //printf("\n");

·                            k = len;

·                            ans = 1;

·                            if(len%(len-next[len])==0)

·                                  ans = len / (len-next[len]);

·                            printf("%d\n", ans);

·                      }

·

·                      return 0;

·                }