POJ 2406 Power Strings （KMP next数组应用）

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 30862 Accepted: 12864

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.

解决这道题的关键，得先学这个定理：

定理：假设S的长度为len，则S存在循环子串，当且仅当，len可以被len - next[len]整除，最短循环子串为S[len - next[len]]

例子证明：
设S=q₁q₂q₃q₄q₅q₆q₇q₈，并设next[8] = 6，此时str = S[len - next[len]] = q₁q₂，由字符串特征向量next的定义可知，q₁q₂q₃q₄q₅q₆ = q₃q₄q₅q₆q₇q₈，即有q₁q₂＝q₃q₄，q₃q₄＝q₅q₆，q₅q₆＝q₇q₈，即q₁q₂为循环子串，且易知为最短循环子串。由以上过程可知，若len可以被len - next[len]整除，则S存在循环子串，否则不存在。

解法：利用KMP算法求出next数组，然后判断：如果串的长度len可以被你len-next [len]整除，

      输出len/(len-next[len]);否则，输出1。

AC代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>#define INF 0x7fffffffusing namespace std;int next[1000005];void getnext(string p,int *next) //next数组的求法{    int m=p.size();    int i=0,j=-1;    next[0]=-1;    while(i<m)    {        if(j==-1 || p[i]==p[j])        {            i++;            j++;            next[i]=p[i]!=p[j]?j:next[j];        }        else  j=next[j];    }}int main(){    string s;    while(cin>>s && s[0]!='.')    {        memset(next,0,sizeof(next));        int l=s.size();        getnext(s,next);        if(l%(l-next[l])==0)  cout<<l/(l-next[l])<<endl;        else  cout<<1<<endl;    }    return 0;}