(POJ2635)The Embarrassed Cryptographer(大数取模)

The Embarrassed Cryptographer

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 13041 Accepted: 3516

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

Nordic 2005

判断所给的数是不是存在小于L的质因子;

思路:大数取模,

#include <set>
#include <map>
#include <list>
#include <stack>
#include <cmath>
#include <queue>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define PI cos(-1.0)
#define RR freopen("input.txt","r",stdin)
using namespace std;
typedef long long LL;
const int MAX = 1e6+100;
int Arr[100000];
int top;
bool vis[MAX];
char str[110];
int b[40];
int L;
bool flag;
int main()
{
    memset(vis,false,sizeof(vis));
    top=0;
    for(LL i=2;i<MAX;i++)
    {
        if(!vis[i])
        {
            Arr[top++]=i;
            for(LL j=i*i;j<MAX;j+=i)
            {
                vis[j]=true;
            }
        }
    }
    while(scanf("%s %d",str,&L))
    {
        if(str[0]=='0'&&L==0)
        {
            break;
        }
        flag=false;
        int len=strlen(str);
        int l=0;
        for(int i=len-1;i>=0;i-=3)
        {
            int ans=0;
            for(int j=2;j>=0;j--)
            {
                if(i-j<0)
                {
                    continue;
                }
                ans=(ans*10+str[i-j]-'0');
            }
            b[l++]=ans;
        }
        int ans;
        for(int i=0;Arr[i]<L;i++)
        {
            ans=0;
            for(int j=l-1;j>=0;j--)
            {
                ans=(ans*1000+b[j])%Arr[i];
            }
            if(ans==0)
            {
                cout<<"BAD "<<Arr[i]<<endl;
                flag=true;
                break;
            }
        }
        if(!flag)
        {
            cout<<"GOOD"<<endl;
        }
    }
    return 0;
}