PAT Grade [September 2019] --A1160 Forever exam [20]

7-1 Forever (20 minutes)

"Forever number" is a positive integer A with K digits, satisfying the following constrains:

the sum of all the digits of A is m;
the sum of all the digits of A+1 is n; and
the greatest common divisor of m and n is a prime number which is greater than 2.

Now you are supposed to find these forever numbers.

Input Specification

Each input file contains one test case. For each test case, the first line contains a positive integer

Output Specification

For each pair of K and m, first print in a line Case X, where X is the case index (starts from 1). Then print n and A in the following line. The numbers must be separated by a space. If the solution is not unique, output in the ascending order of n. If still not unique, output in the ascending order of A. If there is no solution, output No Solution.

Sample Input

2
6 45
7 80

Sample Output

Case 1
10 189999
10 279999
10 369999
10 459999
10 549999
10 639999
10 729999
10 819999
10 909999
Case 2
No Solution

【statement】

　　Since this question has not been on the official website PAT exam, there is no test set is only through a sample, if errors are found, correct me thank comments.

Solution：

　　This question can be violent, but it must be judged in the process, thereby reducing cycle times.

　　DFS is the opponent violence in general, so this question is relatively simple to use DFS

　　But the point is to timely pruning, or no distinction between violent and

　　In particular there is described code comments.

　　[Suddenly found a law, the condition is always after k-2 bits are bits is 9,0-1 and for the m- 9 * (k-2) in combination, if this rule was established, it would be much simpler ^ ^ ...]

 1 #include <iostream>
 2 #include <vector>
 3 #include <string>
 4 #include <algorithm>
 5 using namespace std;
 6 int nums, k, m, n;
 7 vector<pair<int, int>>res;
 8 int divisor(int a, int b)//求最大公约数
 9 {
10     if (a%b == 0)
11         return b;
12     return divisor(b, a %B);
 13 is  }
 14  BOOL the isPrime ( int X) // judgment is not a prime number greater than 
15  {
 16      IF (X <= 2 ) return  to false ;
 . 17      for ( int I = 2 ; I * I <= X; ++ I)
 18 is          IF (% I X == 0 ) return  to false ;
 . 19      return  to true ;
 20 is  }
 21 is  int GetSum ( int X)
 22 is  {
 23 is      int sum = 0;
24     while (x)
25     {
26         sum += x % 10;
27         x /= 10;
28     }
29     return sum;
30 }
31 bool check(int a)
32 {    
33     int b = a + 1;
34     n = 0;
35     while (b)
36     {
37         n += b % 10;
38         B / = 10 ;
 39      }
 40      IF (the isPrime (divisor (m, n-)))
 41 is          return  to true ;
 42 is      the else 
43 is          return  to false ;
 44 is  }
 45  void the DFS ( String & STR, int index, int digit for, int SUM)
 46 is  {
 47      IF (index> SUM = k ||> m) return ; // no more than k bits, starting from 0 
48      IF (+ SUM . 9 * (k - index - . 1 ) <m)return ; // first index number is too small to fill all of the other bits and which are lower than 9 m 
49      STR [index] = digit for + ' 0 ' ; // this step preceding write Oh, because the sum has been added 
50      IF (SUM m && index == K == - . 1 ) // satisfy 
51 is      {
 52 is          int a = Stoi (str.c_str ());
 53 is          IF (Check (a))
 54 is              res.push_back ( the make_pair (n-, A));
 55          return ;
 56 is      }
 57 is      for ( int I = 0 ; I < 10 ; ++i)
58         DFS(str, index + 1, i, sum + i);    
59 }
60 int main()
61 {    
62     cin >> nums;
63     for (int i = 1; i <= nums; ++i)
64     {
65         cin >> k >> m;
66         res.clear();
67         string str(k, '0');
68         for (int j = 1; j < 10; ++j)
69             DFS(str, 0, j, j);//第一位不能为0
70         sort(res.begin(), res.end(), [](pair<int, int>v1, pair<int, int>v2) {//排序
71             if (v1.first == v2.first)
72                 return v1.second < v2.second;
73             return v1.first < v2.first;
74             });
75         printf("Case %d\n", i);
76         if (res.size() > 0)
77             for (auto a : res)
78                 cout << a.first << " " << a.second << endl;
79         else
80             printf("No Solution\n");
81     }
82     return 0;
83 }