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The Embarrassed Cryptographer

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4062		Accepted: 804

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 <= 10¹⁰⁰ and 2 <= L <= 10⁶. 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

Sample Output

GOOD
BAD 11
GOOD
BAD 23
GOOD
BAD 31

題目的大意是：給一個很大的數K，和一個普通的整數L，問K有沒有小于L的質因子，有則輸出“BAD 那個因子”，否則輸出“GOOD”。

首先，明顯當然要打一個素數表了。
接下來就是關鍵部分了，讀入K，把K轉成千進制。把數字往大進制轉換能夠加快運算效率。千進制的性質與十進制相似。例如，1234567890轉成千進制，就變成了：[1][234][567][890]。

然后再從小到大枚舉每一個素數，并對其進行高精度求余就行了。

下面是關鍵部分的一些代碼：

int divide(int div) //高精度求余。 
{
    int i,ans=0;
    for(i=la-1;i>=0;i--)
        ans=(ans*1000+a[i])%div;
    return ans;    
}

void makeprime() //建立素數表。 
{
    int i,j,isprime,k,p=3;
    for(i=6;i<N;i++)
    {
        isprime=1;
        k=(int)sqrt(i);    
        for(j=0;j<p;j++)
        {
            if(prime[j]>k+1)
                break;
            if(i%prime[j]==0)
            {
                isprime=0;
                break;
            }    
        }
        if(isprime) prime[p++]=i;    
    }    
}

        len=strlen(s);
        for(i=0;i<len;i++) //轉化為千進制。 
        {
            t=(len-i+2)/3-1;
            a[t]=a[t]*10+s[i]-'0';
        }
        la=(len+2)/3;