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            posts - 74,  comments - 33,  trackbacks - 0
            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

            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
             
            題目的大意是:給一個(gè)很大的數(shù)K,和一個(gè)普通的整數(shù)L,問K有沒有小于L的質(zhì)因子,有則輸出“BAD 那個(gè)因子”,否則輸出“GOOD”。
            首先,明顯當(dāng)然要打一個(gè)素?cái)?shù)表了。
            接下來就是關(guān)鍵部分了,讀入K,把K轉(zhuǎn)成千進(jìn)制。把數(shù)字往大進(jìn)制轉(zhuǎn)換能夠加快運(yùn)算效率。千進(jìn)制的性質(zhì)與十進(jìn)制相似。例如,1234567890轉(zhuǎn)成千進(jìn)制,就變成了:[1][234][567][890]。

            然后再從小到大枚舉每一個(gè)素?cái)?shù),并對(duì)其進(jìn)行高精度求余就行了。
            下面是關(guān)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?cái)?shù)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(zhuǎn)化為千進(jìn)制。 
                    {
                        t=(len-i+2)/3-1;
                        a[t]=a[t]*10+s[i]-'0';
                    }
                    la=(len+2)/3;
            posted on 2008-12-24 21:51 KNIGHT 閱讀(333) 評(píng)論(0)  編輯 收藏 引用

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