Blue Bridge cup mixed number (full array)

 

Description

100 may be represented in the form of mixed fractions: 69 258 + 3 = 100/714.
It can also be expressed as: 82 + 100 = 3546/197.
Note wherein: a mixed fraction, the numbers 1 to 9 appears once and only once, respectively (not including 0).
Mixed fraction like this, there are 11 100 notation.

Input

From the standard input a positive integer N (N <1000 * 1000)

Output

With a digital program outputs the digital 1 ~ 9 will not be repeated without omission mixed fraction consisting of all the number of species represented.
Note: does not require the output of each said only statistics on the number representation!

Sample Input 1

100

Sample Output 1

11

 

Sample Input 2

105

Sample Output 2

6

 

 

 

 

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <string>
 5 #include <math.h>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <queue>
 9 #include <set>
10 #include <stack>
11 #include <map>
12 #include <sstream>
13 const int INF=0x3f3f3f3f;
14 typedef long long LL;
15 const  int MOD = 1E9 + . 7 ;
 16  const  int MAXN = 1E7 + 10 ;
 . 17  the using  namespace STD;
 18 is  
. 19  int A [ 10 ]; // store full permutation sequence 
20 is  
21 is  int main ()
 22 is  {
 23 is      int n-;
 24      Scanf ( " D% " , & n-);
 25      for ( int I = 0 ; I < . 9 ; I ++) // initialization 
26         A [I] = I + . 1 ;
 27      int ANS = 0 ;
 28      do {
 29          int A = 0 ; // integer part 
30          for ( int I = 0 ; I <= . 6 ; I ++) // integer part up to six 
31 is          {
 32              A * = A 10 + A [I];
 33 is              int B = 0 ; // molecular moiety 
34 is              int C = 0 ; // denominator portion 
35              IF(= A> n-)
 36                  BREAK ;
 37 [              for ( int J = I + . 1 ; J <= I + ( . 9 - (I + . 1 ) + . 1 ) / 2 ; J ++) // molecular moiety least the remaining digit half 
38 is                  B = B * 10 + A [J];
 39              for ( int J = I + ( . 9 - (I + . 1 ) + . 1 ) / 2 + . 1 ; J < . 9 ; J ++) // remaining bits all to the denominator 
40                  C = * C 10 + A [J];
41 is              IF (B% C == 0 && A + B / C == n-) // understood numerator / denominator must be an integer 
42 is                  ANS ++ ;
 43 is              for ( int J = I + ( . 9 - (I + . 1 ) + . 1 ) / 2 + . 1 ; J < . 8 ; J ++) // molecular moiety becomes long, denominator becomes short 
44 is              {
 45                  B = B * 10 + A [J];
 46 is                  C = cA [J] * POW ( 10 , . 9 - (J + . 1 ));
 47                  IF (B% C == 0&&a+b/c==n)
48                     ans++;
49             }
50         }
51     }while(next_permutation(A,A+9));
52     printf("%d\n",ans);
53     return 0;
54 }