joj 1873 Power Strings
1873: Power Strings
Result | TIME Limit | MEMORY Limit | Run Times | AC Times | JUDGE |
---|---|---|---|---|---|
10s | 8192K | 650 | 258 | Standard |
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).
Each test case is a line of input representing s, a string of printable characters. For each s you should print the largest n such that s = a^n for some string a. 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.
Sample Input
abcd aaaa ababab .
Output for Sample Input
1 4 3
This problem is used for contest: 73
Submit / Problem List / Status / Discuss
一个基础的周期串问题。
有更好的办法,不过还是很黄很暴力的枚举吧(大爱各种暴力流水流)...
不过这样写真心跑得很慢...
1 #include <stdio.h> 2 #include <string.h> 3 4 int main() 5 { 6 char s[1000005]; 7 int n; 8 int i, j; 9 10 //freopen("in.txt", "r", stdin); 11 12 while (scanf("%s", s) == 1, s[0] != '.') 13 { 14 n = strlen(s); 15 16 for (i=1; i<=n; ++i) 17 { 18 if (0 == n % i) 19 { 20 int ok = 1; 21 for (j=i; j<n; ++j) 22 { 23 if (s[j] != s[j%i]) 24 { 25 ok = 0; 26 break; 27 } 28 } 29 if (ok) 30 { 31 printf("%d\n", n/i); 32 break; 33 } 34 } 35 } 36 } 37 38 39 return 0; 40 }