Power Strings[poj2406]题解

Power Strings

Description

- Given two strings a and b we define ab to be their concatenation. For example, if a = "abc" and b = "def" then ab = "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 1

- abcd
  aaaa
  ababab
  .

Sample Output 1

- 1
  4
  3

思路1

暴力

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;

const int Max=1000001;
string a;
int x,n,m;
int p[Max];

void prime(int n)
{
    int k=(int)sqrt(n*1.0);
    p[1]=1;
    for(int i=2; i<=k; i++)
        if(n%i==0)
            p[++m]=i;
    int end=m;
    if(k*k==n)  end--;
    for(int i=end; i>=1; i--)
        p[++m]=n/p[i];
}

bool ok(int x)
{
    string s=a.substr(0,x);
    for(int i=0; i<a.size(); i+=x)
        if(s!=a.substr(i,x))    return false;
    return true;
}

int main()
{
    while(getline(cin,a))
    {
        x=1,n=a.size();m=1;
        prime(n);
        while(x<=m)
        {
            if(ok(p[x]))
            {cout<<a.size()/p[x]<<endl;break;}
            x++;
        }
    }
    return 0;
}

思路2

利用\(KMP\)的\(next[]\)数组

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;

const int Max=1000001;
string a;
int N,M,ans,nx[Max];

void makenx(int M)
{
    memset(nx,0,sizeof(nx));
    int i=0,j=-1;
    nx[i]=j;
    while(i<M)
    {
        if(j==-1||a[i]==a[j])   i++,j++,nx[i]=j;
        else    j=nx[j];
    }
}

int main()
{
    bool fl;
    while(getline(cin,a)&&a[0]!='.')
    {
        fl=true;
        M=a.size();makenx(M);
        if(M%(M-nx[M])==0)  cout<<M/(M-nx[M])<<endl;
        else    cout<<1<<endl;
    }
    return 0;
}