Power Strings
Time Limit: 3000MS
Memory Limit: 65536K
Total Submissions: 33178
Accepted: 13792
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
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
Waterloo local 2002.07.01
用next数组求出整个数组的最大前缀。假设整个串是用循环节组成的，那么 n - next[n] 也就是最小循环节，验证最小循环节会被n整出。
#include
#include
#include
int next[1100000] ;
char str[1100000] ;
void getnext(int l)
{
int j = 0 , k = -1 ;
next[0] = -1 ;
while(j < l)
{
if( k == -1 || str[j] == str[k] )
{
j++ ;
k++ ;
next[j] = k ;
}
else
k = next[k] ;
}
}
int main()
{
int l , m ;
while(scanf("%s", str)!=EOF)
{
if( str[0] == '.' ) break;
l = strlen(str);
getnext(l) ;
m = next[l];
if( l % (l-m) != 0 )
printf("1\n");
else
{
m = l / ( l-m );
printf("%d\n", m);
}
memset(str,0,sizeof(str));
}
return 0;
}
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