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  • Poj 2406--Power Strings(KMP)

    Power Strings
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 38544   Accepted: 16001

    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.

    Source

     
    KMP,   扩展数组 ;
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    using namespace std;
    char str[1000010]; int len, NeXt[1000010];
    void GetP()
    {
        int i = 0, j = -1;
        NeXt[i] = j;
        while(i < len)
        {
            if(j == -1 || str[i] == str[j])
            {
                i++; j++;
                if(str[i] != str[j])
                    NeXt[i] = j;
                else
                    NeXt[i] = NeXt[j];
            }
            else
                j = NeXt[j];
        }
            
    } 
    int main()
    {
        while(gets(str), str[0] != '.')
        {
            
            len = strlen(str);
            GetP();
            if(len%(len-NeXt[len])==0)
                printf("%d
    ", len/(len-NeXt[len]));
            else
                printf("1
    ");
        }
        return 0;    
    } 
     
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  • 原文地址:https://www.cnblogs.com/soTired/p/4823239.html
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