xtu字符串 B. Power Strings

B. Power Strings

Time Limit: 3000ms

Memory Limit: 65536KB

64-bit integer IO format:  %lld      Java class name: Main

 

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.

 

解题：求字符串的循环节长度。利用KMP的适配数组。如果字符长度可以被（字符长度-fail[字符长度])整除，循环节这是这个商，否则循环节长度为1，即就是这个字符本身。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cstdlib>

 5 #include <vector>

 6 #include <climits>

 7 #include <algorithm>

 8 #include <cmath>

 9 #define LL long long

10 #define INF 0x3f3f3f

11 using namespace std;

12 const int maxn = 1000100;

13 char str[maxn];

14 int fail[maxn];

15 void getFail(int &len) {

16     int i,j;

17     len = strlen(str);

18     fail[0] = fail[1];

19     for(i = 1; i < len; i++) {

20         j = fail[i];

21         while(j && str[j] != str[i]) j = fail[j];

22         fail[i+1] = str[j] == str[i] ? j+1:0;

23     }

24 }

25 int main() {

26     int len;

27     while(gets(str) && str[0] != '.') {

28         getFail(len);

29         if(len%(len-fail[len])) puts("1");

30         else printf("%d\n",len/(len-fail[len]));

31     }

32     return 0;

33 }

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