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poj2406 kmp 求最小循环字串

2024-09-14 11:07:29   218人阅读

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Power Strings
 
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 47748   Accepted: 19902

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.
 
题意：
给你一个字符串，求该字符串最多由多少个循环字串构成。
 
题解：
　　定义一个整型变量 m;   //m表示字符串长度
　　用kmp里的next数组 求出字符串的最长前缀长度，然后判断 m 是否能被 m-next[m] 整除，即 m%(m-next[m])是否等于0，如果能被整除，m-next[m]为最短循环字串，否则最短循环字串长度等于整个字符串长度。
 
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int next[1000002];
char a[1000002];

void getNext()
{
    int i=-1,j=0;
    next[0]=-1;
    int cnt=0;
    while(a[j])
    {
        if(a[i]==a[j]||i==-1)
        {
            next[++j]=++i;
        }
        else
            i=next[i];
    }
}
int main()
{
    while(scanf("%s",a)!=EOF)
    {
        if(a[0]==‘.‘) break;
        int m=strlen(a);
        getNext();
        int cc=1;
        if(m%(m-next[m])==0)  //如果条件满足，m-next[m]为最短循环字串，然后求出该字符串由多少个最短循环字串构成
            cc=m/(m-next[m]);
        cout<<cc<<endl;
    }
    return 0;
}

　　

poj2406 kmp 求最小循环字串


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