poj2635 The Embarrassed Cryptographer

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.

首先用线性筛求出1e6以内的质数，然后用所有不超过l的质数验证。
这样问题就转化成了大整数取模。逐位加上数，边加边取模即可。

#include<cstdio>#include<cstring>const int mx=1e6;int prm[1000010],tot;bool have[1000010];char s[110];void make(){    for (int i=2;i<=mx;i++)    {        if (!have[i]) prm[++tot]=i;        for (int j=1;j<=tot&&i*prm[j]<=mx;j++)        {            have[i*prm[j]]=1;            if (i%prm[j]==0) break;        }    }}int main(){    int i,j,k,l,m,n,p,q,x,y,z;    bool flag;    make();    while (scanf("%s%d",s+1,&l)&&l)    {        m=strlen(s+1);        flag=0;        for (i=1;i<=tot&&prm[i]<l;i++)        {            n=0;            for (j=1;j+2<=m;j+=3)              n=(n*1000+(s[j]-'0')*100+(s[j+1]-'0')*10+s[j+2]-'0')%prm[i];            if (j==m-1) n=(n*100+(s[j]-'0')*10+s[j+1]-'0')%prm[i];            if (j==m) n=(n*10+s[j]-'0')%prm[i];            if (n==0)            {                flag=1;                printf("BAD %d\n",prm[i]);                break;            }        }        if (!flag) printf("GOOD\n");    }}