poj 2635

The Embarrassed Cryptographer

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12076		Accepted: 3224

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
K是两个质数的积，L是个范围，若能找到一个小于L的质数是K的约数，那么BAD同时输出该质数，否则GOOD
题目很明朗了，就是要怎么处理大数了，自己看代码吧，理解。
AC代码：

#include<iostream>
#include<string>
#include<stdio.h>
#include<cstring>
using namespace std;
string s;
int prime[500100];
int isprime[1000100];
int num[110];
int lenp,L;
void sieve(int n){    //筛素数
int i,j;
for(i=0;i<=n;i++)
isprime[i]=1;
lenp=0;
for(i=2;i<=n;i++){
if(isprime[i]){
prime[lenp++]=i;
for(j=2*i;j<=n;j+=i)
isprime[j]=0;
}
}
}
int main(){
sieve(1000000);
while(cin>>s>>L){
if(s[0]=='0' && L==0)
break;

int l=s.length();            //大数转化为数字，每三位存在num里
int a=l/3;
int b=l%3;
int i,j,k;
for(i=0;i<=a;i++)
num[i]=0;
if(b){
for(i=0;i<b;i++)
num[0]=num[0]*10+s[i]-'0';
}
j=b;
for(i=1;i<=a;i++){
for(k=0;k<3 && j<l;j++,k++){
num[i]=num[i]*10+s[j]-'0';
}
}
int sucess=0;
int ans;
for(j=0;j<lenp && prime[j]<L;j++){
int ans=0;
for(i=0;i<=a;i++){
ans=(ans*1000+num[i])%prime[j];
}
if(ans==0){                              //判断能否被整除
sucess=1;
k=j;
break;
}
}
if(sucess)
cout<<"BAD "<<prime[k]<<endl;
else
cout<<"GOOD"<<endl;
}
return 0;
}