The Embarrassed Cryptographer

最新推荐文章于 2019-04-03 15:22:09 发布

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本文介绍了一个关于素数筛选与高精度求模的问题背景及解决方案。在一个使用两个素数乘积作为密钥的安全系统中，为确保密钥强度，通过筛选素数表并运用同余模定理来验证密钥的有效性。

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The Embarrassed Cryptographer

Describe
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

Sample Output

GOOD
BAD 11
GOOD
BAD 23
GOOD
BAD 31

高精度求模，先筛出一个素数表（注意素数表需要含有一个大于一百万的数，如果输入的L是一百万，而一百万内没有K的素因数，按照题目要求也输出GOOD，会造成程序一直运行导致数组越界而RE），然后根据同余模定理对K一步步求模。

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int prime[100000];
int flag[1000010]={1,1};

int main()
{
    int top=0;
    for(int i=2;i<=1000010;i++)
    {
        if(!flag[i])
        {
            prime[top++]=i;
            for(int j=i*2;j<=1000010;j+=i)
            {
                flag[j]=1;
            }
        }
    }
    int i,j,k;
    char s[200];
    while(scanf("%s %d",s,&k)!=EOF)
    {
        if(strcmp(s,"0")==0&&k==0)
            break;
        int len=strlen(s);
        for(i=0;i<top;i++)
        {
            int rest=0;
            for(j=0;j<len;j++)
            {
                rest=(rest*10+s[j]-'0')%prime[i];
            }
            if(rest==0)
                break;
        }
        if(i!=top&&prime[i]<k)
            printf("BAD %d\n",prime[i]);
        else
            printf("GOOD\n");
    }
    return 0;
}