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.
这题不就是和上一个题类似,还简单点,直接就是(len)/n,len为字符串长度,n为循环节的长度,当len%n==0
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1e6+10;
char s[maxn];
int ne[maxn];
void get(int len)
{
ne[0]=-1;
int j=-1;
for(int i=0;i<len;)
{
if(j==-1||s[i]==s[j] )ne[++i]=++j;
else j=ne[j];
}
}
int main()
{
while(~scanf("%s",s)&&s[0]!='.')
{
int len=strlen(s);
get(len);
int n=len-ne[len];//循环节的长度
if(len%n==0) //中间没有杂数
printf("%d\n",len/n);
else printf("1\n");
}
}
/*
len=18;
a b a b c a b a b a b a b c a b a b
-1 0 0 1 0 0 1 2 3 0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 1
*/