HDOJ 題目2303 The Embarrassed Cryptographer（數學）

The Embarrassed Cryptographer

Time 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 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

Source NCPC2005
Recommend zty | We have carefully selected several similar problems for you: 2300 2308 2301 2305 2306 輸入一個大數和一個整數k，然後看看那個大大數能不能整除一個素數，要是可以看看那個素數比k大還是小，要是小就輸出BAD，並把那個素數輸出出來，否則輸出GOOD ac代碼

#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