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POJ 2406 Power Strings

时间:2016-05-04 22:46:42      阅读:306      评论:0      收藏:0      [点我收藏+]
F - Power Strings
Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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.
 
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<map>
 5 #define maxn 10010000
 6 using namespace std;
 7 char p[maxn];
 8 int next[maxn],len;
 9 int kmp(){
10     int i=0,j=-1;
11     next[0]=-1;
12     while(i<len){
13         if(j==-1||p[i]==p[j]) next[++i]=++j;
14         else j=next[j];
15     }
16 }
17 int main()
18 {
19     while(scanf("%s",p)!=EOF&&p[0]!=.){
20         len=strlen(p);
21         kmp();
22         if(len%(len-next[len]))printf("1\n");
23         else printf("%d\n",len/(len-next[len]));
24     }
25     return 0;
26 }

 

POJ 2406 Power Strings

原文:http://www.cnblogs.com/shenben/p/5459926.html

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