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.
nxt数组典型应用求循环节
对于长度n 循环节次数就是n/(n-nxt[n]) 很容易推导
#include<cstdio>
#include<string.h>
using namespace std;
char s[400005];
int nxt[400005];
int n;
void get_nxt()
{
int i=0;
int k=-1;
nxt[0]=-1;
while(i<n)
{
if(k==-1||s[i]==s[k])
{
i++;
k++;
nxt[i]=k;
}
else
{
k=nxt[k];
}
}
return;
}
int ans[400005];
int main()
{
while(~scanf("%s",s))
{
n=strlen(s);
get_nxt();
int cnt=0;
ans[++cnt]=n;
int x=n;
while(x)
{
x=nxt[x];
if(x!=0)
{
ans[++cnt]=x;
}
}
for(int i=cnt;i>=1;i--)
{
printf("%d%s",ans[i],i==1?"
":" ");
}
}
return 0;
}