【POJ2406】 Power Strings （KMP）

Power Strings

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.

 

 

【题意】

　　一个字符串的最大重复字串。

　　即一个字符串是一个子串最多重复多少次得到的。

 

【分析】

　　

　　① len==1||s[next[len-1]]!=s[len-1] ans=1；
　　② len%(len-next[len])==0 ans= len/(len-next[len]);

　　原因如图：

　　

代码如下：

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<queue>
 using namespace std;
 #define Maxn 1000010

 char s[Maxn];
 int len,nt[Maxn];

 void kmp()
 {
     nt[]=;
     int p=;
     for(int i=;i<=len;i++)
     {
         while(s[i]!=s[p+]&&p) p=nt[p];
         if(s[i]==s[p+]) p++;
         nt[i]=p;
     }
 }

 int main()
 {
     int n,i;
     while()
     {
         gets(s+);
         len=strlen(s+);
         if(len==&&s[]=='.') break;
         kmp();
         if(len%(len-nt[len])==) printf("%d\n",len/(len-nt[len]));
         else printf("1\n");
     }
 }

[POJ2406]

2016-07-20 16:30:02