E - The Embarrassed Cryptographer POJ - 2635

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

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

取模优化，将%的运算次数减少可提高程序的效率，另外感觉自己这个素数筛的真够猥琐的，要去学一学了。

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
using namespace std;
char str[105];
int primer[80000],l,len;
bool prim[1000005];
bool isprim(int x){
int t = sqrt(x);
for(int i = 2;i <= t;++i){
if(x % i == 0)
return false;
}
return true;
}
void init(){
len = 0;
for(int i = 2;i <= 1000;i++){
if(!prim[i]){
for(int j = i << 1;j <= 1000000;j += i){
prim[j] = true;
}
}
}
for(int i = 2;i <= 1000000;++i){
if(!prim[i]){
primer[len++] = i;
}
}
//  cout << len << endl;
}
bool isdiv(int x){
long long ans = 0;

b1c7
for(int i = 0;str[i] != '\0';++i){
ans = ans * 10 + str[i] - '0';
if(ans >= 100000000000000000)
ans = ans % x;
}
ans = ans % x;
if(ans == 0){
return true;
}
else{
return false;
}
}
int main(){
init();
while(~scanf("%s%d",str,&l)){
if(str[0] == '0' && l == 0){
break;
}
bool flag = false;
int ans = 0;
for(int i = 0;i < len;++i){
if(primer[i] >= l){
break;
}
//cout << primer[i] << endl;
if(isdiv(primer[i])){
ans = primer[i];
flag = true;
break;
}
}
if(flag){
printf("BAD %d\n",ans);
}
else{
printf("GOOD\n");
}
}
return 0;
}