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POJ 2406:Power Strings

言午栩 2022-08-10 阅读 94


Power Strings


Time Limit: 3000MS

 

Memory Limit: 65536K

Total Submissions: 41252

 

Accepted: 17152


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.



题意:

给出的一个字符串某个字串链接N次得出的,求最大N;


思想:

KMP匹配算法之next[];模式串结尾指针回溯最小移动次数即最小字串长度,N=s.length%(s.length-next[s.length]);



#include <stdio.h>
#include <string.h>
char s[1000005];
int num[1000005];
int main()
{
while(scanf("%s",s),s[0]!='.')
{
int j=-1;
num[0]=-1;
int n=strlen(s);
for(int i=0; i<n;)
{
if(j==-1||s[i]==s[j])num[++i]=++j;
else j=num[j];
}
if(n%(n-num[n])==0)
printf("%d\n",n/(n-num[n]));
else printf("1\n");
}
return 0;
}



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