hdu 2303 The Embarrassed Cryptographer

The Embarrassed Cryptographer

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 377    Accepted Submission(s): 109

Problem Description

The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of users, which is now in use in his company. The cryptographic keys are created from the product of two primes, and are believed to be secure because there is no known method for factoring such a product effectively. 
What Odd Even did not think of, was that both factors in a key should be large, not just their product. It is now possible that some of the users of the system have weak keys. In a desperate attempt not to be fired, Odd Even secretly goes through all the users keys, to check if they are strong enough. He uses his very poweful Atari, and is especially careful when checking his boss' key.

 

Input

The input consists of no more than 20 test cases. Each test case is a line with the integers 4 <= K <= 10 ¹⁰⁰ and 2 <= L <= 10 ⁶. K is the key itself, a product of two primes. L is the wanted minimum size of the factors in the key. The input set is terminated by a case where K = 0 and L = 0.

 

Output

For each number K, if one of its factors are strictly less than the required L, your program should output "BAD p", where p is the smallest factor in K. Otherwise, it should output "GOOD". Cases should be separated by a line-break.

 

Sample Input

  
  

   
   143 10
143 20
667 20
667 30
2573 30
2573 40
0 0
  
  

 

Sample Output

  
  

   
   GOOD
BAD 11
GOOD
BAD 23
GOOD
BAD 31
  
  
  
  


  
  
  
  

   
   将十进制数化为千进制，然后大整数取模。
  
  
  
  

   
   AC代码：
  
  
  
  

   
   
#include<iostream>
#include<string>
#include<cstring>
using namespace std;
bool prime[1000001];
int len1,len2;

int in[40];

void build_prime()                   //筛选素数
{
    memset(prime,1,sizeof(prime));
    prime[0]=prime[1]=0;
    for(int i=2; i<1000; i++)
    {
        for(int j=i; i*j<1000001; j++)
            prime[i*j]=0;

    }
}

void change(string temp)       //十进制转化为千进制
{
    memset(in,0,sizeof(in));
    int i,j;
    len1=0;
    int t=temp.length()%3;
    for(i=0; i<t; i++)
        in[len1]=in[len1]*10+(temp[i]-'0');
    if(t)
        len1++;
    for(i=t; i<temp.length(); i++)
    {
        for(j=1; j<=3&&i<temp.length(); j++)
        {
            in[len1]=in[len1]*10+(temp[i++]-'0');
        }
        len1++;
        i--;
    }
}
bool division(int pri)             //大整数取模
{
    int i;
    int t=0;
    for(i=0; i<len1; i++)
    {
        t*=1000;
        t+=in[i];
        t%=pri;
    }
    if(t==0)
        return true;

    return false;
}
int main()
{
    build_prime();
    string temp;
    int L,i,flag=1;
    while(cin>>temp>>L&&temp!=""&&L)
    {
        flag=1;
        change(temp);
        if(L>2&&division(2))
        {
            cout<<"BAD "<<2<<endl;
            flag=0;
        }
        else
        {
            for(i=3; i<L; i+=2)
            {
                if(prime[i])
                {
                    if(division(i))
                    {
                        cout<<"BAD "<<i<<endl;
                        flag=0;
                        break;
                    }

                }
            }
        }

        if(flag)
            cout<<"GOOD"<<endl;
    }
    return 0;
}