POJ 2635 The Embarrassed Cryptographer(素数筛+高精度求模+同余模)

The Embarrassed Cryptographer

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12906		Accepted: 3473

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

给一个非常大的数K，和一个L(2<=L<=10^6)

其中K 是两个素数的乘积，如果这两个素数中的最小值比L大 输出"GOOD" , 否则输出"BAD" + 两个素数中小的那个

可以先将L内的素数全部存起来，来依次判断是否 K%某素数=0

因为K非常大，所以可以把它化成千进制，如12345678 化为 [12][345][678]

判断[12][345][678]%5 是否为0

先判断 12%5=2；

下一步 (345+2*1000)%5=0;

最后  （678+0*1000）%5=3；

所以 12345678 不能整除 5。

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#define N 1000001
using namespace std;
int su[N],s;
bool u[N];
void sai()
{
    memset(u,true,sizeof(u));
     s=1;
    for(int i=2; i<N; i++)
    {
        if(u[i]) su[s++]=i;
        for(int j=1; j<s; j++)
        {
            if(i*su[j]>N) break;
              u[i*su[j]]=false;
            if(i%su[j]==0) break;
        }
    }
}
int a[N];
int main()
{
    char str[10010];
    int n,p;
    sai();                  //素数筛
    while(scanf("%s%d",str,&n),n)
    {
        int len=strlen(str);
        int mm=len%3;
        int aa=0;
        if(mm==1)                   //换成千进制
           a[aa++] = str[0]-'0';
        else if(mm==2)
         a[aa++] = (int)(str[0]-'0')*10 + (int)(str[1]-'0');
        for(int i=mm;i<len;i+=3)
          a[aa++] = (str[i]-'0')*100 + (str[i+1]-'0')*10 + str[i+2]-'0';
        
        bool flag=false;
        int mul=1000;
        for(int i=1;i<s;i++)         //依次判断每个素数
        {
            if(su[i]>n) break;
            int sum=a[0]%su[i];
            for(int j=1;j<aa;j++)    //是否能整除
            {
                sum = (a[j]+mul*sum)%su[i];
            }
            if(sum==0)
            {
                if(su[i]<n)
                {
                    p=su[i];
                    flag=true;
                    break;
                }
            }
        }
        if(flag)
          printf("BAD %d\n",p);
          else printf("GOOD\n");
    }
    return 0;
}