编程之美2.8 | 找符合条件的整数

思路还是相当地巧妙。

求余数的话，(a+b)%n=(a%n+b%n)%n;

用vector来表示整数的话（出现1的位置），可以避免溢出。

注意第20行，在更新remainders[(j+r)%n]时，要确保每个remainders的每个序列都是递增的，不能存在相等的情况。

 1 #include <time.h>

 2 #include <math.h>

 3 #include <stdlib.h>

 4 

 5 using namespace std;

 6 

 7 long long findMInZeroOneNum(int n) {

 8         vector<vector<int> > remainders(n);

 9         remainders[1].push_back(0);

10 

11         bool update = false;

12         int noupdate = 0;

13         for (int i = 1, j = 10 % n; ; j = (j * 10) % n, i++) {

14                 if (remainders[j].empty()) {

15                         remainders[j].push_back(i);

16                         update = true;

17                 }

18                 for (int r = 1; r < n; ++r) {

19                         if (remainders[r].empty()) continue;

20                         if (i <= remainders[r].back()) continue;

21                         if (!remainders[(j + r) % n].empty()) continue;

22                         remainders[(j + r) % n] = remainders[r];

23                         remainders[(j + r) % n].push_back(i);

24                         update = true;

25                 }

26                 if (!update) noupdate++;

27                 else noupdate = 0;

28                 if (noupdate == n || !remainders[0].empty()) break;

29         }

30 

31         if (remainders[0].empty()) return -1;

32         else {

33                 long long ans = 0;

34                 for (int i = 0; i < remainders[0].size(); ++i) {

35                         ans += pow(10, remainders[0][i]);

36                 }

37                 return ans;

38         }

39 }

40 

41 long long bruteForce(int n) {

42         if (n <= 0) return -1;

43         for (long long i = 1; ; ++i) {

44                 if (i % n != 0) continue;

45                 long long tmp = i;

46                 while (tmp) {

47                         int d = tmp % 10; 

48                         if (d != 0 && d != 1) break;

49                         tmp /= 10;

50                 }

51                 if (tmp == 0) return i;

52         }

53         return -1;

54 }

55 

56 int main() {

57         srand(time(NULL));

58         int n = rand() % 99;

59 

60         cout << n << endl;

61         long long n1 = findMInZeroOneNum(n);

62         cout << n1 << endl;

63         long long n2 = bruteForce(n);

64         cout << n2 << endl;

65         if (n1 != n2) {

66                 cout << n << " " << n1 << " " << n2 << endl;

67         }

68 

69         return 0;

70 }