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poj 2406 Power Strings
Power Strings
http://poj.org/problem?id=2406
| Time Limit: 3000MS | Memory Limit: 65536K | |
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.
求最短循环节的出现次数
求最短循环节长度的方法:
a=len-next[len]
若len%a==0 ，最短循环节长度为a
否则，没有循环节
#include<cstdio>#include<cstring>#include<iostream>using namespace std;int n;char s[1000001];int f[1000010];void getnext(){ for(int i=1;i<n;i++) { int j=f[i]; while(j&&s[i]!=s[j]) j=f[j]; f[i+1]= s[i]==s[j] ? j+1:0; }}int main(){ while(cin>>s) { if(s[0]==‘.‘) break; n=strlen(s); getnext(); int ans=n-f[n]; if(n%ans) printf("1\n"); else printf("%d\n",n/ans); } }
poj 2406 Power Strings
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