Power Strings
Time Limit:3000MSMemory Limit:65536K
Total Submissions:34716Accepted:14367
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
abcdaaaaababab.
Sample Output
143
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
循环节 ，，求字符串的循环的最大数量。
#include<stdio.h>#include<iostream>#include<math.h>#include<stdlib.h>#include<ctype.h>#include<algorithm>#include<vector>#include<string.h>#include<queue>#include<stack>#include<set>#include<map>using namespace std;char b[1000500];int Next[1000500];void get_next(char b[], int m){int i = 0,j = -1;memset(Next,0,sizeof(Next));Next[0] = -1;while (b[i]){if (j == -1 || b[i] == b[j]){++i;++j;Next[i] = j;}elsej = Next[j];}}int main(){int cases = 1, n, m, i, j, ans;while (scanf("%s",&b)!=EOF && b[0]!='.'){n = strlen(b);get_next(b, n);int t = Next[n];int s = n – t;if (n % s == 0)ans = n / s;elseans = 1;printf("%d\n",ans);}return 0;}
但我想说，我做了一个善良的平凡女子，并且一直在爱，
