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：自己看（最后遇到“.”要break）
解：
len为整个字符串的长度
想一想next[i]的含义
每次匹配失败后，都要回到next[i]开始重新匹配
也就是说 next[i]+1 -> len 这段字符 (令k为这段字符的长度)和
1 -> 1+k 这段字符 是相同的
OK，如果(len%(len-next[len])==0) ==> 存在 len/(len-next[len]) 段 解
否则只有1段
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #include<string> 7 #include<queue> 8 #include<map> 9 using namespace std; 10 const int N=1e7; 11 char s[N]; 12 int len,ne[N],j; 13 int main() 14 { 15 while(scanf("%s",s+1)!=EOF) 16 { 17 if(s[1]==‘.‘) break; 18 len=strlen(s+1); 19 for(int i=0;i<=len;++i) ne[i]=0; 20 for(int i=2;i<=len;++i) 21 { 22 j=ne[i-1]; 23 while(j && s[i]!=s[j+1]) j=ne[j]; 24 if(s[i]==s[j+1]) ne[i]=j+1; 25 } 26 if(len%(len-ne[len])==0) printf("%d\n",len/(len-ne[len])); 27 else printf("1\n"); 28 } 29 return 0; 30 }
