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  • Power Strings

    Total Submission(s) : 10 Accepted Submission(s) : 4
    Problem 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
    #include<iostream>
    #include<string.h>
    using namespace std;
    int next[1000005],len;
    void set_naxt(char str[])
    {
        int i=0,j=-1;
        next[0]=-1;
        len=strlen(str);
        while(i<len)
        {
            if(j==-1||str[i]==str[j])
            {
                i++; j++;
                next[i]=j;
            }
            else
            j=next[j];
        }
    }
    int main()
    {
        int I;
        char str[1000001];
        while(cin>>str&&strcmp(str,".")!=0)
        {
            set_naxt(str);
            if(len%(len-next[len])==0)
            I=len/(len-next[len]);
            else
            I=1;
            cout<<I<<endl;
        }
    }
    /*len=strlen(str);
            for(i=1;i<=len;i++)
            {
                for(j=i;j<len;j++)
                if(str[j]!=str[j%i])
                break;
                if(j==len)
                break;
            }
    */
    


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  • 原文地址:https://www.cnblogs.com/jiangu66/p/3220308.html
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