poj2405 Power Strings KMP
Power Strings
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.
思路:
KMP預處理nxt陣列存前面的最長前後綴,對於整個字串來說,如果len % (len - nxt[len - 1]) == 0,那麼這個字串肯定是有周期的並且週期T = len / (len - nxt[len - 1]),於是我們只需要暴力遍歷一遍整個字串就行了;
程式碼:
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn = (int)1e6 + 10; char s[maxn]; int nxt[maxn]; void init(int n) { memset(nxt,0,sizeof(nxt)); nxt[0] = 0; int i = 1,j = 0; while (i < n) { if (s[i] == s[j]) nxt[i ++] = ++ j; else if (!j) i ++; else j = nxt[j - 1]; } } int main() { while (~scanf("%s",s)) { if (!strcmp(s,".")) break; int len = strlen(s); init(len); if (len % (len - nxt[len - 1]) == 0) printf("%d\n",len / (len - nxt[len - 1])); else printf("1\n"); } return 0; }