POJ 2635 The Embarrassed Cryptographer 大数模

题目： http://poj.org/problem?id=2635

利用同余模定理大数拆分取模，但是耗时，需要转化为高进制，这样位数少，循环少，这里转化为1000进制的，如果转化为10000进制，需要long long

 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
 #include <cmath>
 using namespace std; int p[];
 bool prime[];
 void prime_init()
 {
     memset(prime, , sizeof(prime));
     prime[] = ;
     for(int i = ; i <= ; i += )
         prime[i] = ;
     for(int i = ; i <= sqrt(); i += )
     {
         if(prime[i])
         {
             for(int j = i*i; j <= ; j += i+i)
                 prime[j] = ;
         }
     }
     int cnt = ;
     for(int i = ; i < ; i++)
         if(prime[i])p[cnt++] = i;
 } int main()
 {
     char str[];
     int x, num[];
     prime_init();
     while(scanf("%s %d", str, &x) != EOF)
     {
         if(!strcmp(str, "") && x == )break;
         int len = strlen(str);
         int tmp = , k = ;
         for(int i = ; i < len%; i++)
             tmp = tmp *  + str[i] - '';
         num[k++] = tmp;
         for(int i = len%; i < len; i += )
         {
             tmp = ;
             for(int j = i; j < i+; j++)
                 tmp = tmp *  + str[j] - '';
             num[k++] = tmp;
         }
         bool flag = ;
         for(int i = ; p[i] < x && !flag; i++)
         {
             tmp = ;
             for(int j = ; j < k; j++)
                 tmp = (tmp *  + num[j]) % p[i];
             if(tmp == )
             {
                 flag = ;
                 printf("%s %d\n", "BAD", p[i]);
             }
         }
         if(!flag)printf("GOOD\n");
     }
     return ;
 }