POJ 2406 Power Strings KMP 求最小循环节

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 29027 Accepted: 12124

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

Waterloo local 2002.07.01

给定一个字符串m，求一个最短的子串n，m是n重复多次组成的（求m的最小循环节）。

利用kmp中next（我的数组名是p）数组的性质。

表达能力拙计。。。分析样例的next数组找数学规律吧。

还有,用cin是可以AC的。

#include <iostream>using namespace std;string s;int l,i,j,p[1000010];int main(){ios::sync_with_stdio(false);while (cin>>s && s!="."){l=s.size();j=-1; p[0]=-1;for (i=1;i<l;++i){while (j>-1 && s[i]!=s[j+1]) j=p[j];if (s[i]==s[j+1]) ++j;p[i]=j;}i=l-p[l-1]-1;if (l%i==0) cout<<l/i<<endl; else cout<<"1\n";}return 0;} 

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