Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 29402 | Accepted: 12296 |
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
两种做法：
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 char a[1100000]; 6 int t; 7 int main() 8 { 9 char x; 10 while(1) 11 { 12 t=0; 13 x=getchar(); 14 if(x=='.')break; 15 while(1) 16 { 17 a[t++]=x; 18 if(x==' ')break; 19 x=getchar(); 20 } 21 int p=1,k=1; 22 t--; 23 for(int i=0; i<t; i++) 24 { 25 if(a[i]==a[i%p]) 26 k++; 27 else 28 { 29 if(p==k)k++; 30 p=k; 31 } 32 } 33 if(t%p==0) 34 printf("%d ",t/p); 35 else printf("1 "); 36 } 37 }
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 char a[1100000]; 6 int t,next[1100000]; 7 void fun() 8 { 9 next[0]=0; 10 int i,j=0; 11 for(i=1;i<t;i++) 12 { 13 while(j>0&&a[i]!=a[j])j=next[j-1]; 14 if(a[i]==a[j])j++; 15 next[i]=j; 16 } 17 } 18 int main() 19 { 20 char x; 21 while(1) 22 { 23 t=0; 24 x=getchar(); 25 if(x=='.')break; 26 while(1) 27 { 28 a[t++]=x; 29 if(x==' ')break; 30 x=getchar(); 31 } 32 t--; 33 fun(); 34 if(t%(t-next[t-1])==0) 35 printf("%d ",t/(t-next[t-1])); 36 else printf("1 "); 37 } 38 }