矿大OJ 1768.Power Strings.

Posted 2021-02-20 yuuuuu422

题目描述

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).

输入

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.

输出

For each s you should print the largest n such that s = a^n for some string a.

样例输入

abcd
aaaa
ababab
.

样例输出

1
4
3


好像弄成英文看起来就变得高级了 其实就是求emmm.一时间我也不好描述..代码如下

 1 #include <iostream>
 2 #include <string>
 3 using namespace std;
 4 int test(string s)
 5 {
 6     char a = s[0];
 7     int locate = s.find(a, 1);
 8     while(locate > 0)
 9     {
10         bool flag = false;
11         int len = s.length();
12         int times =  len/ locate;
13         if (len % locate == 0)
14         {
15             flag = true;
16             for (int i = locate;i < len&&flag==true;i += locate)
17             {
18                 for (int j = 0;j <= locate - 1;j++)
19                 {
20                     if (s[j] != s[i + j])
21                     {
22                         flag = false;
23                         break;
24                     }
25                 }
26 
27             }
28             
29         }    
30         if(flag)
31             return times;    
32         else
33             locate = s.find(a, locate + 1);
34     }
35     return 1;
36 }
37 
38 int main()
39 {
40     string s;
41     while (cin >> s && s != ".")
42     {
43         cout << test(s) << endl;
44     }
45     return 0;
46 }

这个做了我差不多5个小时..据说好像用什么KMP一些就做出来，现在还没学也看不懂，等到时候接触了再说吧