POJ The Embarrassed Cryptographer 2635 (数论)

　　　　　　　　　　　　　　　　　　　　　The Embarrassed Cryptographer

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14151		Accepted: 3868

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

题意：给出一个大数K，K是两个素数的乘积

            再给出一个小于等于100W的L

            问K的两个素数因子较小的那一个是否小于L

            小于输出这个数，否则输出GOOD (如果在100W内没有找到因子，也输出GOOD)

分析：首先我们先把100W的素数求出来（打表素数筛）

　　　然后根据同余模定理，比如　123对3取余

　　　就是1%3=1

            (1*10+2)%3=0

            (0*10+3)%3=0

            最后的结果就是0

　　　这里需要把K转为千进制的，因为十进制的一位一位的取余会超时

　　　假设 K = 1234567891，转成千进制的就是 [ 1 ] [ 123 ] [ 567 ] [ 891 ] ，用个数组存每一位就行

　　　这样把上面的乘10改成乘1000就可以了

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int Max=1100000;
int pre[Max+10],pr[Max+10],top;
void paly_table()　//素数筛
{
int i,j,t;
memset(pre,0,sizeof(pre));
pre[2]=1;
for(i=3;i<=Max;i++)
{
if(i&1)
pre[i]=1;
}
t=sqrt(Max);
for(i=0;i<=t;i++)
{
if(pre[i])
{
for(j=i*i;j<=Max;j+=2*i)
{
pre[j]=0;
}
}
}
top=0;
for(i=0;i<=1000010;i++)
{
if(pre[i])
{
pr[top++]=i;
}
}
//    for(i=0;i<10;i++)
//    {
//       printf("%d ",pr[top-1]);
//    }
}

int main()
{
char s[200];
int a[200],l,mod,n,m,i,j;
paly_table();
while(~scanf("%s",s))
{
scanf("%d",&mod);
if(s[0]=='0' && mod==0)
break;
l=strlen(s);
n=0;
memset(a,0,sizeof(a));
m=l%3;
if(m==1)　　　//转千进制，用的方法好笨
{
a[0]=s[0]-'0';
n++;
}
else if(m==2)
{
a[0]=(s[0]-'0')*10+(s[1]-'0');
n++;
}
for(i=m;i<l;)
{
int ct=0;
while(ct<3)
{
a
=a
*10+(s[i]-'0');
i++;ct++;
}
n++;
}
//        for(i=0;i<n;i++)
//        {
//            printf("%d ",a[i]);
//        }
//        printf("\n");
int k=0;
int flag=0;
for(i=0;i<top;i++)
{
k=0;
for(j=0;j<=n-1;j++)
{
k=(k*1000+a[j])%pr[i];　//取余
//printf("%d\n",k);
}
if(k==0)
{
flag=1;
m=pr[i];
break;
}
}

if(flag && m<mod)　　//能找到因子，且比给的数小
{
printf("BAD %d\n",m);
}
else
{
printf("GOOD\n");
}

}
return 0;
}