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

 1 #include <stdio.h>

 2 #include <stdlib.h>

 3 #include <string.h>

 4 #include <cmath>

 5 using namespace std;

 6 

 7 int p[78600];

 8 bool prime[1001001];

 9 void prime_init()

10 {

11     memset(prime, 1, sizeof(prime));

12     prime[1] = 0;

13     for(int i = 4; i <= 1001000; i += 2)

14         prime[i] = 0;

15     for(int i = 3; i <= sqrt(1001000); i += 2)

16     {

17         if(prime[i])

18         {

19             for(int j = i*i; j <= 1001000; j += i+i)

20                 prime[j] = 0;

21         }

22     }

23     int cnt = 0;

24     for(int i = 2; i < 1001000; i++)

25         if(prime[i])p[cnt++] = i;

26 }

27 

28 int main()

29 {

30     char str[110];

31     int x, num[10010];

32     prime_init();

33     while(scanf("%s %d", str, &x) != EOF)

34     {

35         if(!strcmp(str, "0") && x == 0)break;

36         int len = strlen(str);

37         int tmp = 0, k = 0;

38         for(int i = 0; i < len%3; i++)

39             tmp = tmp * 10 + str[i] - '0';

40         num[k++] = tmp;

41         for(int i = len%3; i < len; i += 3)

42         {

43             tmp = 0;

44             for(int j = i; j < i+3; j++)

45                 tmp = tmp * 10 + str[j] - '0';

46             num[k++] = tmp;

47         }

48         bool flag = 0;

49         for(int i = 0; p[i] < x && !flag; i++)

50         {

51             tmp = 0;

52             for(int j = 0; j < k; j++)

53                 tmp = (tmp * 1000 + num[j]) % p[i];

54             if(tmp == 0)

55             {

56                 flag = 1;

57                 printf("%s %d\n", "BAD", p[i]);

58             }

59         }

60         if(!flag)printf("GOOD\n");

61     }

62     return 0;

63 }

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