链接:
http://poj.org/problem?id=2406
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.
代码:
#include<stdio.h>
#include<string.h>
#include<stdlib.h> #define N 1000007 char S[N];
int Next[N]; /// Next中存的是前缀和后缀的最大相似度 void FindNext(int Slen, int Next[]) ///Next[i] 代表前 i 个字符的最大匹配度
{
int i=, j=-;
Next[] = -; while(i<Slen)
{
if(j==- || S[i]==S[j])
Next[++i] = ++j;
else
j = Next[j];
}
} int main()
{
while(scanf("%s", S), strcmp(S, "."))
{
int Slen=strlen(S); FindNext(Slen, Next); if(Slen%(Slen-Next[Slen]))
printf("1\n");
else
printf("%d\n", Slen/(Slen-Next[Slen]));
} return ;
}