HDOJ question 2303 The Embarrassed Cryptographer (Mathematics)

HDOJ question 2303 The Embarrassed Cryptographer (Mathematics)
The Embarrassed CryptographerTime Limit: 3000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 563 Accepted Submission (s): 172


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 has tively.
What Odd Even did not think of, was that both factors in a key shocould 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 shoshould output "BAD p", where p is the smallest factor in K. otherwise, it shoshould output "GOOD ". cases shoshould be separated by a line-break.
Sample Input

143 10143 20667 20667 302573 302573 400 0

Sample Output

GOODBAD 11GOODBAD 23GOODBAD 31

Source NCPC2005
Recommend zty | We have carefully selected several similar problems for you: 2300 2308 2301 2305 2306 enter a large number and an integer k, and then check whether the large number can divide a prime number, if you can see whether the prime number is larger or smaller than k, if it is smaller, it will output BAD, and the prime number will be output; otherwise, the GOOD ac code will be output.

#include
 
  #include
  
   int is[1000010],prim[10000100],num=0,k;void fun(){int i,j;for(i=2;i<1000010;i++){if(!is[i]){prim[num++]=i;for(j=i+i;j<1000010;j+=i){is[j]=1;}}}}char s[220];int main(){fun();while(scanf("%s %d",s,&k)!=EOF){if(strcmp(s,"0")==0&&k==0)break;int len=strlen(s),i,j,sum;for(i=0;i