POJ2406----Power Strings解题报告

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 43514		Accepted: 18153

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

1
4
3

Hint

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

 

题目链接：http://poj.org/problem?id=2406

KMP算法，next表示模式串如果第i位(设str[0]为第0位)与文本串第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。对于一个串，如果abcdabc， 那么next[len]=3,那么len-next【len】就大于len/2,那么len%(len-next[len])!=0;而对于一个周期串ababab next[len]=4,此时len-next[len]应该等于最小串的长度，最小周期就可以用len%(len-next[len])是否为0来判断。

 

 

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <string>
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef long double ld;
 const int maxn = ;
 char str[maxn];
 int Next[maxn];
 int len;
 void get_next()
 {
     int j = -;
     int i = ;
     Next[] = ;
     while(i < len)
     {
         if(str[i] == str[j] || j == -)
         {
             i++;
             j++;
             if(str[i] != str[j])
                 Next[i] = j;
             else
                 Next[i] = Next[j];
         }
         else
             j = Next[j];
     }
 }
 
 int main()
 {
     while(~scanf("%s",str))
     {
         if(str[] == '.')
             break;
         len =strlen(str);
         get_next();
         int ans = ;
         if(len%(len-Next[len]) == )
             ans = len/(len-Next[len]);
         //for(int g = 0; g <= len; g++)
         //{
         //    cout << Next[g] << ' ';
         //}
         //cout << endl;
         cout << ans << endl;
     }
     return ;
 }