POJ-2635 The Embarrassed Cryptographer

最新推荐文章于 2019-07-27 11:55:12 发布

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最新推荐文章于 2019-07-27 11:55:12 发布

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分类专栏：数学题易错文章标签： ACM 易错数学题

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本文探讨了在大型系统中使用弱密钥的风险，并详细解释了如何通过高效算法进行检查，确保用户密钥的安全性。文章还提供了一个案例研究，展示了如何通过检查系统中的密钥来识别并修复潜在的弱点。

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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 <= 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

Source

Nordic 2005

题目大意：给你一个高精度数n和一个单精度数l，问n的第一个质因数是否小于l。

分析： WA了好久才过，直接暴力会超时，要用到压位高精度。

#include <cstdio>
#include <cstring>
#include <iostream> 
#define MAXN 1000000
using namespace std;
int num,l,prim[MAXN+1],f[101];
bool notprim[MAXN+1];
char s[101];
int Size(int x)
{
	if(x == 0) return 1000;
	if(x == 2) return 100;
	return 10;
}
int main()
{
	for(int i = 2;i <= MAXN;i++)
	 if(!notprim[i])
	 {
	 	prim[++num] = i;
	 	for(int j = 2*i;j <= MAXN;j += i)
	 	 notprim[j] = true;
	 }
	while(scanf("%s%d",s,&l) && l)
	{	
		int length = strlen(s),tot = 0; 
		bool flag = false;
		for(int i = 0;i < length;i++)
		{
			if(i % 3 == 0) f[++tot] = s[i] - '0';
			else f[tot] = f[tot]*10 + s[i] - '0';
		}
		for(int i = 1;i <= num && prim[i] < l;i++)
		{ 
			int now = 0;
			for(int j = 1;j < tot;j++) 
		 	 now = (now*1000 + f[j]) % prim[i];
		 	now = (now*Size(length % 3) + f[tot]) % prim[i];
		 	if(now == 0) 
		 	{
		 		cout<<"BAD "<<prim[i]<<endl;
		 		flag = true;
		 		break;	
			}
		}
		if(!flag) cout<<"GOOD"<<endl;
	} 
}