Power Strings
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 55320 | Accepted: 22983 |
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.
题解:题意就是由一个字符串重复多遍组成,求最多重复次数。
也就是求这个字符串的最小长度即可。很明显与KMP的fail数组有关,求出可以发现如果重复次数大于1,那么n-fail[n]必定是最小的字符串长度。随便列几次就可以发现了。
Code:
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; #define N 1000005 int next[N],len; int main() { char a[N]; while(~scanf("%s",a)) { if(a[0]=='.')break; len=strlen(a); int i=0,j=-1; next[0]=-1; while(i<len) { if(j==-1||a[i]==a[j]) { i++,j++; next[i]=j; }else j=next[j]; } int k=len-next[len]; if(len%k==0)printf("%d\n",len/k); else printf("1\n"); } return 0; }