POJ2406 Power Strings

 

Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 42637   Accepted: 17787

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.

Source

 

KMP求出next数组,设字符串长度为len,若len/(len-next[len])为整数,那么这个数就是答案。

类似的题见http://www.cnblogs.com/SilverNebula/p/5550595.html

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 int next[1000010];
 8 char s[1000010];
 9 int len,ans;
10 void getnext(){
11     int i,j;
12     next[0]=0;
13     next[1]=0;
14     for(i=1,j=0;i<len;i++){
15         while(s[j]!=s[i] && j)j=next[j];
16         if(s[j]==s[i])j++;
17         next[i+1]=j;
18     }
19     return;
20 }
21 int main(){
22     while(scanf("%s",s)!=EOF){
23         if(s[0]=='.')break;
24         len=strlen(s);
25 //        memset(next,0,sizeof next);
26         getnext();
27         ans=0;
28         if(len%(len-next[len])==0)
29             printf("%d\n",len/(len-next[len]));
30         else printf("1\n");
31     }
32     return 0;
33 }

 

posted @ 2016-07-17 09:07  SilverNebula  阅读(195)  评论(0编辑  收藏  举报
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