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。
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cstring>
5 #include <string>
6 using namespace std;
7 typedef long long ll;
8 typedef unsigned long long ull;
9 typedef long double ld;
10 const int maxn = 1000010;
11 char str[maxn];
12 int Next[maxn];
13 int len;
14 void get_next()
15 {
16 int j = -1;
17 int i = 0;
18 Next[0] = 0;
19 while(i < len)
20 {
21 if(str[i] == str[j] || j == -1)
22 {
23 i++;
24 j++;
25 if(str[i] != str[j])
26 Next[i] = j;
27 else
28 Next[i] = Next[j];
29 }
30 else
31 j = Next[j];
32 }
33 }
34
35 int main()
36 {
37 while(~scanf("%s",str))
38 {
39 if(str[0] == ‘.‘)
40 break;
41 len =strlen(str);
42 get_next();
43 int ans = 1;
44 if(len%(len-Next[len]) == 0)
45 ans = len/(len-Next[len]);
46 //for(int g = 0; g <= len; g++)
47 //{
48 // cout << Next[g] << ‘ ‘;
49 //}
50 //cout << endl;
51 cout << ans << endl;
52 }
53 return 0;
54 }