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poj2406 Power Strings
Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 33273 | Accepted: 13825 |
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
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
Waterloo local 2002.07.01
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 int next[1000010],m; 8 char s[1000010]; 9 10 void get_next()11 {12 int i=0,j=-1;13 next[0]=-1;14 while(i<m)15 {16 if(j==-1||s[i]==s[j])17 {18 i++;19 j++;20 next[i]=j;21 }22 else j=next[j];23 }24 }25 int main()26 {27 while(scanf("%s",s)!=EOF)28 {29 if(s[0]==‘.‘)break;30 m=strlen(s);31 get_next();32 if(m%(m-next[m])==0)printf("%d\n",m/(m-next[m]));33 else printf("1\n");34 }35 return 0;36 }
poj2406 Power Strings
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