sicily 1231. The Embarrassed Cryptography

1231. The Embarrassed Cryptography

Constraints

Time Limit: 2 secs, Memory Limit: 32 MB

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 10143 20667 20667 302573 302573 400 0

Sample Output

GOODBAD 11GOODBAD 23GOODBAD 31

题目分析

给定一个大整数和num,看大整数能否被比num小的素数整除
大整数除法,利用同余定理

#include <stdio.h>#include <memory.h>int prime[80000];int count;void init() {  int visited[1000005];  memset(visited, true, sizeof(visited));  for (int c = 2; c < 1000005; ++c)    if (visited[c])      for (int d = c + c; d < 1000005; d += c)        visited[d] = false;  count = 0;  for (int c = 2; c < 1000005; ++c)    if (visited[c])      prime[count++] = c;}int main(){  init();  char key[101];  int test;  while (scanf("%s%d", key, &test)) {    if (key[0] == '0' && test == 0)      break;    int hope = 0;    for (int c = 0; prime[c] < test; ++c) {      int temp = 0;      for (int d = 0; d < strlen(key); ++d) {        temp = (temp * 10) + (key[d] - '0');        temp %= prime[c];      }      if (temp == 0) {        hope = prime[c];        break;      }    }    if (hope)      printf("BAD %d\n", hope);    else      printf("GOOD\n");  }  return 0;}