Description
Given two strings a and b we define ab to be their concatenation. For example, if a = “abc” and b = “def” then ab = “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
题意
给出的一个字符串是某个字串连接N次得出的,求最大N的。
思路
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]);
AC 代码
#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;
}