素数表

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 

 

#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>

using namespace std;

int kw[1001];
bool zhengchu(int n,int lkw)					//是否能整除？
{
	int sum=0;
	for(int i=lkw-1;i>=0;i--)
		sum=(sum*1000+kw[i])%n;

   if(sum%n==0)
	   return true;
   return false;
}

bool sushu[1000001]={0};//0为素数

void ss()
{
	sushu[0]=1;
	sushu[1]=1;
	for(int i=2;i*i<1000005;i++)
	{
		if(sushu[i]==0)
		{
			for(int j=2*i;j<1000005;j+=i)
				sushu[j]=1;
		}
	}
}
int main()
{
	string s;
	int n;
	cin>>s;
	ss();
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
			break;

	  memset(kw,0,sizeof(kw));
	  int sum=0;
	  int i=0;
	 for(i=0;i<s.length();i++)
	 {
	     int ii=(s.length()+2-i)/3-1;
		  kw[ii]=s[i]-'0'+10*kw[ii];
	 }
		for(i=2;i<n;i++)
		{
			if(!sushu[i]&&zhengchu(i,(s.length()+2)/3))
			{
				cout<<"BAD"<<' '<<i<<endl;
				break;
			}
		}
		if(i==n)
			cout<<"GOOD"<<endl;

		cin>>s;
	}


}

从这道题，可以学到

判断素数的简便方法

bool sushu[1000001] = { 0 };//0为素数

void ss()

{

sushu[0] = 1;

sushu[1] = 1;

for (int i = 2; i*i<1000005; i++)

{

if (sushu[i] == 0)

{

for (int j = 2 * i; j<1000005; j += i)

sushu[j] = 1;

}

}

}

 

还有大数据的进制转换，将大数转化为千进制数据，加快处理，

Int kw[10000]；

对s输入的字符串进行进制转换，

 memset(kw,0,sizeof(kw));

  int i=0;

 for(i=0;i<s.length();i++)

 {

     int ii=(s.length()+2-i)/3-1;

  kw[ii]=s[i]-'0'+10*kw[ii];

 }

注意高位在kw的后面，首先处理，然后把处理剩下的数据*1000+前一位。

int kw[1001];

bool zhengchu(int n,int lkw)

{

//是否能整除？

int sum=0;

for(int i=lkw-1;i>=0;i--)

sum=(sum*1000+kw[i])%n;

   if(sum%n==0)

   return true;

   return false;

}