poj 2635 The Embarrassed Cryptographer(素数筛 + 同余模定理)

本文介绍了一种算法，用于检查大数值是否由足够大的质因数构成，以确保加密密钥的安全性。通过将大数转换为千进制并使用预计算的质数表进行验证。

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The Embarrassed Cryptographer

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14106		Accepted: 3858

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

Sample Output

GOOD
BAD 11
GOOD
BAD 23
GOOD
BAD 31
参考思路
解题思路：给定两个数，一个数S是由两个素数 a,b相乘得到的数，另一个数L随便，问 min(a,b) 是否小于L。
解题思路：素数打表，然后从前往后遍历看S是否能被当前素数整除，如果整除，判断是否小于L。
注意：给定的S是大数，只能用字符数组保存。
一个数a判断能不能整除b，只要判断 a%b是否等于0就可以了。
同余定理：
(a+b)%c=(a%c+b%c)%c;
(a*b)%c=(a%c+b%c)%c;
对大数取余
模板：  大数字符串形式 a[1000];
char a[1000];
   int m=0;
   for(int i=0;a[i]!='\0';i++)
       m=((m*10)%n+(a[i]-'0')%n)%n;//也可以写成  m=(m*10+a[i]-'0')%n
m为所求的余数
本题大数求余的方法为： 把字符串从后面起 三位三位的保存在一个int类型数组中，比如 12345   存在int类型数组里面为  12   345  然后按照前面的方法求：
bool mod(int n)
{
    int m=0;
    for(int i=0;i<=k-1;i++)
       m=(m*1000+num[i])%n;
    if(m==0)
        return true;
    return false;
}
//另一个
<p>求一个10^100 的K 对 小于 10^6的 L 取余，问2-L-1 之间能否有能整除K的素数。</p><p>设k=a0a1a2a3a4a5a6a7a8a9a10.</p><p>k % m =a0a1a2a3a4a5a6*1000 %m + a7a8a9a10%m</p><p>          =(a0a1a2*1000 %m + a3a4a5a6)*1000 %m + a7a8a9a10%m</p><p>于是很明显可以递归求解，k%m</p><p>于是枚举m，每次求解k就好了。。。。</p><p>注意素数打表的时候用j=i+i！！！</p><p>//</p><p>一开始害怕K太大存不下来而没有去细算直接转化为万进制，结果一直WA，后来转化为千进制就过了，</p><p>原因是 for循环时每次取余的时候，余数最大可达10^6,然后再乘10000就会爆int，然后又改成了 long long，转化为万进制，也是可以过的
</p>
<pre name="code" class="cpp">#include<stdio.h>
#include<string.h>

#define maxn 1000005
char s[1100];
int isprime[maxn];
int prime[maxn];
int num[maxn];
int lenprime;
int len;
int k;
void play_table()
{
    int i,j;
    for(i=1;i<=maxn;i++)
        isprime[i]=1;

    isprime[0]=0;
    isprime[1]=0;
    lenprime = 0;
    for(i = 2; i <=maxn; i++)
    {
        if(isprime[i])
        {
            prime[lenprime++]=i;
            for(j=i*2;j<=maxn;j+=i)
                isprime[j] = 0;
        }
    }
}
void change()
{
     k = 0;
    int a = len/3;
    int b = len%3;
    int temp = 0;
    int i,j;
    if(b)
    {
        for(i=0;i<b;i++)
        temp=(temp*10+s[i]-'0');
        num[k++] = temp;
    }


    for(i=b; i <len; i+=3)
    {
        num[k++] = (s[i]-'0')*100+(s[i+1]-'0')*10+s[i+2]-'0';
    }

}
int mod(int n)
{
    int i,j;
    int m = 0;
    for(i = 0; i <= k-1; i++)
    {
        m=(m*1000+num[i])%n;
    }
    if(m==0)
        return 1;
    else
        return 0;
}
int main()
{

    int l;
    int i,j;
    int ok;
    int ans;
    play_table();
    while(~scanf("%s%d",s,&l))
    {
        ok  = 0;
        len = strlen(s);
        if(l+s[0]-'0'==0) break;
        change();
        for(i=0;i<lenprime;i++)
        {
            if(mod(prime[i])&&prime[i]<l)
            {
                ok = 1;
                ans = prime[i];
                break;
            }
        }

        if(!ok)
            printf("GOOD\n");
        else
            printf("BAD %d\n",ans);
    }
return 0;
}