POJ 2406-Power Strings（重复子串-KMP中的next数组）

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Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 47642 Accepted: 19867

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

题目意思：

一个字符串由K个循环节组成，求循环节最小的时候其字符的个数L。

解题思路：

暴力又又TLE了，然后，发现要利用next数组的特性：模式串长度为n，其第1位到next[n]与模式串第n-next[n]位到n位是匹配的。

根据定义，next[i]表示模式串0~i-1长度中，后缀与前缀的最长匹配子串的长度，所以next[n]表示整个模式串后缀与前缀的最长匹配子串的长度，

next[KL] = (K-1)L，即next[n] = n - L，所以L = n-next[n]，所以n-next[n]表示一个最小循环节的长度。

题解参考

#include<iostream>#include<cstdio>#include<iomanip>#include<cmath>#include<cstdlib>#include<cstring>#include<map>#include<algorithm>#include<vector>#include<queue>using namespace std;#define INF 0xfffffff#define MAXN 1000010char t[MAXN];int next[MAXN];int main(){#ifdef ONLINE_JUDGE#else    freopen("G:/cbx/read.txt","r",stdin);    //freopen("G:/cbx/out.txt","w",stdout);#endif    ios::sync_with_stdio(false);    cin.tie(0);    while(cin>>t)    {        if(t[0]=='.') break;        memset(next,0,sizeof(next));        int len=strlen(t);        int ans=1;        int p=0,cur;        next[0]=-1;        next[1]=0;        for(cur=2; cur<=len; ++cur)//求next数组        {            while(p>=0&&t[p]!=t[cur-1])                p=next[p];            next[cur]=++p;        }        if(len%(len-next[len])==0)//模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的            ans=len/(len-next[len]);        cout<<ans<<endl;    }    return 0;}

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