POJ2406-Power Strings(kmp循环节)

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 31111		Accepted: 12982

DescriptionGiven 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 = "" (theempty string) and a^(n+1) = a*(a^n).InputEach 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.OutputFor 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

题意：输出循环节最大的个数

思路：next函数推断一下就可以：

若n%(n-next)==0 ans = n/(n-next)

否则 ans = 1

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <string>#include <algorithm>#include <queue>using namespace std;const int maxn = 1000000+10;int next[maxn];char str[maxn];int n;void getNext(){next[0] = next[1] = 0;for(int i = 1,j; i < n; i++){j = next[i];while(j && str[i] != str[j]) j = next[j];if(str[i]==str[j]) next[i+1] = j+1;else next[i+1] = 0;}}int main(){while(~scanf("%s",str) && strcmp(str,".")!=0){n = strlen(str);getNext();int ans;if(n%(n-next)!=0) ans = 1;else ans =  n/(n-next);printf("%d\n",ans);}return 0;}