POJ 2406 Power Strings (KMP)

Power Strings

Time Limit: 3000MS
Memory Limit: 65536K

Total Submissions: 29663
Accepted: 12387

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

abcd
aaaa
ababab
.

Sample Output

1
4
3

::初學KMP真心不容易，理解起來很吃力，還好今天老師有講解KMP的基本原理，現在思路清晰了一點，趕緊做這道以前暴力過的題。

證明：（PKU 2406 POWER STRINGS --- 字符串匹配，KMP算法）

定理：假設S的長度為len，則S存在循環子串，當且僅當，len可以被len - next[len]整除，最短循環子串為S[len - next[len]]
例子證明：

設S=q₁q₂q₃q₄q₅q₆q₇q₈，並設next[8] = 6，此時str = S[len - next[len]] = q₁q₂，由字符串特征向量next的定義可知，q₁q₂q₃q₄q₅q₆ = q₃q₄q₅q₆q₇q₈，即有q₁q₂＝q₃q₄，q₃q₄＝q₅q₆，q₅q₆＝q₇q₈，即q₁q₂為循環子串，且易知為最短循環子串。由以上過程可知，若len可以被len - next[len]整除，則S存在循環子串，否則不存在。
解法：利用KMP算法，求字符串的特征向量next，若len可以被len - next[len]整除，則最大循環次數n為len/(len - next[len])，否則為1。

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1e6;
char s[maxn+10];
int next[maxn+10];

void get_next(char s[],int len)
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j]) {j++; i++; next[i]=j;}
        else j=next[j];
    }
}

int main()
{
    while(scanf("%s",s),s[0]!='.')
    {
        int len=strlen(s);
        get_next(s,len);
        if(len%(len-next[len])==0)
            printf("%d\n",len/(len-next[len]));
        else
            printf("1\n");
    }
    return 0;
}