POJ2406----Power Strings解题报告

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 43514		Accepted: 18153

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.

 

题目链接：http://poj.org/problem?id=2406

 

 KMP算法，next表示模式串如果第i位(设str[0]为第0位)与文本串第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。对于一个串，如果abcdabc， 那么next[len]=3,那么len-next【len】就大于len/2,那么len%(len-next[len])!=0;而对于一个周期串ababab next[len]=4,此时len-next[len]应该等于最小串的长度，最小周期就可以用len%(len-next[len])是否为0来判断。

 

 

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int maxn = 1000010;
char str[maxn];
int Next[maxn];
int len;
void get_next()
{
    int j = -1;
    int i = 0;
    Next[0] = 0;
    while(i < len)
    {
        if(str[i] == str[j] || j == -1)
        {
            i++;
            j++;
            if(str[i] != str[j])
                Next[i] = j;
            else
                Next[i] = Next[j];
        }
        else
            j = Next[j];
    }
}

int main()
{
    while(~scanf("%s",str))
    {
        if(str[0] == '.')
            break;
        len =strlen(str);
        get_next();
        int ans = 1;
        if(len%(len-Next[len]) == 0)
            ans = len/(len-Next[len]);
        //for(int g = 0; g <= len; g++)
        //{
        //    cout << Next[g] << ' ';
        //}
        //cout << endl;
        cout << ans << endl;
    }
    return 0;
}