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Title effect: pressing numbers on the keypad, only on the basis of this position by the right down, and outputs the minimum absolute values ​​of difference to the requirements

AC code is as follows:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <string>
 5 using namespace std;
 6 const int maxn=1e9+10;
 7 int a[10][10];
 8 bool vis[210];
 9 int f[210];
10 int k;
11 
12 bool judge(int x)
13 {
14     int TOT = 0 ;
 15      int S [ 10 ];
 16      for (; X; X / = 10 )
 . 17      {
 18 is          S [TOT ++] = X% 10 ;
 . 19      }
 20 is      for ( int I = TOT; I> . 1 ; i--) // determines whether the subject every given output condition 
21 is      {
 22 is          iF (a [S [I]] [S [I-! . 1 ]]) return  to false ;
 23 is      }
 24      return  to true ;
 25  }
 26 
27  void the init ()
 28  {
 29      // play table 1.A [i] [j] denotes the i key to play key table 1 j, i from the other not to bond to 0 
30      A [ 1 ] [ 1 ] = . 1 ; A [ . 1 ] [ 2 ] = . 1 ; A [ . 1 ] [ . 3 ] = . 1 ; A [ . 1 ] [ . 4 ] = . 1 ; A [ . 1 ] [ . 5 ] = . 1 ; A [ . 1 ] [ . 6 ] = . 1 ; A [ . 1 ] [ . 7 ] = . 1 ; A [ . 1 ] [8]=1; a[1][9]=1; a[1][0]=1;
31     a[2][2]=1; a[2][3]=1; a[2][5]=1; a[2][6]=1; a[2][8]=1; a[2][9]=1; a[2][0]=1;
32     a[3][3]=1; a[3][6]=1; a[3][9]=1;
33     a[4][4]=1; a[4][5]=1; a[4][6]=1; a[4][7]=1; a[4][8]=1; a[4][9]=1; a[4][0]=1;
34     a[5][5]=1; a[5][6]=1; a[5][8]=1; a[5][9]=1; a[5][0]=1;
35     a[6][6]=1; a[6][9]=1;
36     a[7][7]=1; a[7][8]=1; a[7][9]=1; a[7][0]=1;
37     a[8][8]=1; a[8][9]=1; a[8][0]=1;
38     a[9][9]=1;
39     a[0][0 ] = . 1 ;
 40  
41 is      Memset (VIS, to false , the sizeof (VIS));
 42 is      Memset (F, - . 1 , the sizeof (F));
 43 is      // play table 2, all the numerals coincidence output condition between 0-210 the number 
44 is      for ( int i = 0 ; i <= 210 ; i ++ )
 45      {
 46 is          IF (Judge (i)) VIS [i] = to true ;
 47      }
 48      // play table 3, violence test i line with output number of test. 1-I, I-2, I-2,1,0 ,,,. 3; I; I +. 1, I + 2,. 3 I +, I + 210 .... 
49      for ( int i=1; i<=200; i++ )
50     {
51         for(int j=0; j<=210; j++ )
52         {
53             if( i-j>=0 && vis[i-j] )
54             {
55                 f[i]=i-j;
56                 break;
57             }
58             if( i+j<=210 && vis[i+j] )
59             {
60                 f[i]=i+j;
61                 break;
62             }
63         }
64     }
65 }
66 
67 int main()
68 {
69     int t;
70     scanf("%d",&t);
71     init();
72     while( t-- )
73     {
74         scanf("%d",&k);
75         printf("%d\n",f[k]);
76     }
77      Return  0 ;
78 }