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 }