poj2635The Embarrassed Cryptographer（同余膜定理）

The Embarrassed Cryptographer

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15069		Accepted: 4132

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

/*
* @Author: lyucheng
* @Date:   2017-10-17 19:03:06
* @Last Modified by:   lyucheng
* @Last Modified time: 2017-10-17 16:50:37
*/
#include <stdio.h>
#include <vector>
#include <string.h>

#define MAXN 105
#define MAXM 1000005

using namespace std;

char str[MAXN];
int k;
int p[MAXM];
bool prime[MAXM];
int tol;
int num[MAXM];
vector<int>v;

void init(){
    tol=;
    for(int i=;i<MAXM;i++){
        if(prime[i]==false)
            p[tol++]=i;
        for(int j=;j<tol&&i*p[j]<MAXM;j++){
            prime[i*p[j]]=true;
            if(i%p[j]==)
                break;
        }
    }
}

bool ok(int k){
    int s=;
    for(int i=(int)v.size()-;i>=;i--){
        s*=;
        s%=k;
        s+=v[i];
        s%=k;
    }
    if(s==)
        return true;
    else
        return false;
}

int main(){
    // freopen("in.txt","r",stdin);
    init();
    while(scanf("%s%d",str,&k)!=EOF&&(str[]-''!=&&k!=)){
        int n=strlen(str);
        v.clear();
        for(int i=n-;i>=;i-=){
            int s=;
            for(int j=max(,i-);j<=i;j++){
                s*=;
                s+=str[j]-'';
            }
            v.push_back(s);
        }
        // for(int i=0;i<(int)v.size();i++){
        //     cout<<v[i]<<" ";
        // }cout<<endl;
        bool flag=true;
        for(int i=;p[i]<k;i++){
            if(ok(p[i])==true){
                printf("BAD %d\n",p[i]);
                flag=false;
                break;
            }
        }
        if(flag==true)
            puts("GOOD");
    }
    return ;
}