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

Output

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

给定两个数，一个数S是由两个素数 a,b相乘得到的数，另一个数L随便，问 min(a,b) 是否小于L。

解题思路：素数打表，然后从前往后遍历看S是否能被当前素数整除，如果整除，判断是否小于L。

注意：给定的S是大数，只能用字符数组保存。

一个数a判断能不能整除b，只要判断 a%b是否等于0就可以了。

同余定理：

(a+b)%c=(a%c+b%c)%c;

(a*b)%c=(a%c+b%c)%c;

对大数取余

模板：  大数字符串形式 a[1000];

char a[1000];

   int m=0;

   for(int i=0;a[i]!=‘\0‘;i++)

       m=((m*10)%n+(a[i]-‘0‘)%n)%n;//也可以写成  m=(m*10+a[i]-‘0‘)%n

m为所求的余数

本题大数求余的方法为： 把字符串从后面起 三位三位的保存在一个int类型数组中，比如 12345   存在int类型数组里面为  12   345  然后按照前面的方法求：

bool mod(int n)
{
    int m=0;//气死我了！！！！！！
    for(int i=0;i<=k-1;i++)
       m=(m*1000+num[i])%n;
    if(m==0)
        return true;
    return false;
}

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn=1000005;
bool isprime[maxn+1];
int prime[maxn+1];
int primelen=0;//素数表的长度
int slen;//输入大数的长度
char s[10000];int l;
int num[600];int k;//转换后的长度

void sieve(int n)//筛法求素数
{
    for(int i=0;i<=n;i++)
        isprime[i]=1;
    isprime[0]=isprime[1]=0;
    for(int i=2;i<=n;i++)
    {
        if(isprime[i])
        {
            prime[primelen++]=i;
            for(int j=2*i;j<=n;j+=i)
                isprime[j]=0;
        }
    }
}

void change()//将字符串每三位（从后数）放在int数组中，比如 12345 ，  12   345 
{
    k=0;
    int a=slen/3;
    int b=slen%3;
    int temp=0;
    if(b)//不能被3整除
    {
    for(int i=0;i<b;i++)
        temp=temp*10+s[i]-'0';
        num[k++]=temp;
    }
    int cnt=0;
    for(int i=1;i<=a;i++)
    {
        temp=0;
        int start=b+3*cnt;
        for(int j=0;j<3;j++)
            temp=temp*10+s[start+j]-'0';
        num[k++]=temp;
        cnt++;
    }
}

bool mod(int n)
{
    int m=0;//气死我了！！！！！！
    for(int i=0;i<=k-1;i++)
       m=(m*1000+num[i])%n;
    if(m==0)
        return true;
    return false;
}

int main()
{
    sieve(maxn);
    while(cin>>s>>l&&(s[0]!='0'||l!=0))
    {
        slen=strlen(s);
        change();
        bool ok=1;
        int ans;
        for(int i=0;i<primelen;i++)
        {
            if(mod(prime[i])&&prime[i]<l)
            {
                ok=0;
                ans=prime[i];
                break;
            }
            if(prime[i]>=l)
                break;
        }
        if(ok)
            cout<<"GOOD"<<endl;
        else
            cout<<"BAD "<<ans<<endl;
    }
    return 0;
}

[ACM] POJ 2635 The Embarrassed Cryptographer （同余定理，素数打表）