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E - Power Strings,求最小周期串

2024-07-14 03:29:56   253人阅读

E - Power Strings
Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 2406

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

abcdaaaaababab.

Sample Output

143


讲了这么多，其实就是一个求一个字符串里最小周期的问题。限时有3S，我们可以用枚举。
这里需要注意的是一个：求周期串的基本算法：
for(i=1;i<=len;i++)
{
  ok=1;
  if(len%i==0)
    {
     for(j=i;j<len;j++)
      {
        if(s[j]!=s[j%i]){ok=0;break;}
      }
    }
if(ok==1)
{
 printf("%d\n",len/i);
  break;//记得要break
}
}
 1 #include<cstdio> 2 #include<string.h> 3 using namespace std; 4 char s[1000100]; 5 int main() 6 { 7     while(scanf("%s",s)==1&&strcmp(s,".")!=0) 8     { 9         int len=strlen(s);10 11         for(int i=1;i<=len;i++)12             if(len%i==0)13         {14              int ok=1;15             for(int j=i;j<len;j++)16             {17                 if(s[j]!=s[j%i])18                 {19                     ok=0;20                     break;21                 }22             }23             if(ok)24             {25                  printf("%d\n",len/i);26                  break;27             }28 29         }30     }31     return 0;32 }

 


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