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[78600];
bool prime[1001001];
void prime_init()
{
    memset(prime, 1, sizeof(prime));
    prime[1] = 0;
    for(int i = 4; i <= 1001000; i += 2)
        prime[i] = 0;
    for(int i = 3; i <= sqrt(1001000); i += 2)
    {
        if(prime[i])
        {
            for(int j = i*i; j <= 1001000; j += i+i)
                prime[j] = 0;
        }
    }
    int cnt = 0;
    for(int i = 2; i < 1001000; i++)
        if(prime[i])p[cnt++] = i;
}

int main()
{
    char str[110];
    int x, num[10010];
    prime_init();
    while(scanf("%s %d", str, &x) != EOF)
    {
        if(!strcmp(str, "0") && x == 0)break;
        int len = strlen(str);
        int tmp = 0, k = 0;
        for(int i = 0; i < len%3; i++)
            tmp = tmp * 10 + str[i] - '0';
        num[k++] = tmp;
        for(int i = len%3; i < len; i += 3)
        {
            tmp = 0;
            for(int j = i; j < i+3; j++)
                tmp = tmp * 10 + str[j] - '0';
            num[k++] = tmp;
        }
        bool flag = 0;
        for(int i = 0; p[i] < x && !flag; i++)
        {
            tmp = 0;
            for(int j = 0; j < k; j++)
                tmp = (tmp * 1000 + num[j]) % p[i];
            if(tmp == 0)
            {
                flag = 1;
                printf("%s %d
", "BAD", p[i]);
            }
        }
        if(!flag)printf("GOOD
");
    }
    return 0;
}