Triangle Conjecture， ACM-DIY Group Contest 2011 Spring 之 6，HDOJ 3805

Triangle Conjecture， ACM-DIY Group Contest 2011 Spring 之 6，HDOJ 3805

Triangle Conjecture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Special Judge

Problem Description

One could construct a triangle with the digit 1 to 9 as the figure below:

The triangle is the one that the sums of every four numbers on its three edges are all equals to 23. Moreover, 23 is the biggest summation one can get from this kind of arraignment of digits. Your task is even tougher, given a positive integer n, you should use integer from 1 to 3*(n-1) to construct triangle with equal summation of digits on the three edges and the summation is the biggest among all the possible arraignments. For example, if n = 4, then you should choose number from 1 to 3*(4-1).
For convenience, the output format for a certain triangle is like the example for the figured triangle above:
95 41 38 2 6 7
The numbers are separated by a single space in each row, and there are no spaces at the end of each row.
Note that there may be several solutions exist, arbitrary one of them will be accepted.

 

Input

The first line of the input contains a number t indicates the number of test cases.
Following t lines, each line will contains only one integer n denoting the side length of the desired triangle. ( t≤20, 3≤n≤1000)

 

Output

For each test case, output the triangles one by one.

 

Sample Input

   
   
   
   

    
    
    
    2
    
    
    
    
3
    
    
    
    
4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    6
    
    
    
    
1 2
    
    
    
    
5 3 4
    
    
    
    
9
    
    
    
    
5 4
    
    
    
    
1 3
    
    
    
    
8 2 6 7
   
   
   
   

 

Author

ahui

 

Source

ACM-DIY Group Contest 2011 Spring

找规律，n 分奇偶处理，比赛时仓促，代码不很简洁

  1  #include  < stdio.h >
  2 
  3  #define   N  1009
  4 
  5  int  a[ N ], b[ N ], c[ N ], m, n;
  6 
  7  void  solveOdd() {
  8           int  i  =   1 ;
  9          m  =   0 ;
 10           while  ( i  <=   3 * n - 6  ) {
 11                   if  ( i & 1  ) {
 12                          a[ m ]  =  i ++ ;
 13                          b[ m ]  =  i ++ ;
 14                          c[ m ]  =  i ++ ;
 15                  }
 16                   else  {
 17                          c[ m ]  =  i ++ ;
 18                          b[ m ]  =  i ++ ;
 19                          a[ m ]  =  i ++ ;
 20                  }
 21                   ++ m;
 22          }
 23  }
 24 
 25  void  solveEven() {
 26           int  i;
 27           for  ( i  =   1 ; i  <=  n - 2 ;  ++ i ) {
 28                  b[ i - 1  ]  =  n - 2 + i;
 29          }
 30          a[  0  ]  =   2 ;
 31          a[  1  ]  =   3 * n - 6 ;
 32          c[  0  ]  =   1 ;
 33          c[  1  ]  =   3 * n - 7 ;
 34          m  =   2 ;
 35          i  =   2 ;
 36           while  ( m  <  n - 2  ) {
 37                   ++ i;
 38                  a[ m ]  =  i;
 39                  a[ m + 1  ]  =   3 * n - 5 - i;
 40                   ++ i;
 41                  c[ m ]  =  i;
 42                  c[ m + 1  ]  =   3 * n - 5 - i;
 43                  m  +=   2 ;
 44          }
 45  }
 46 
 47  void  swap(  int  a[],  int  b[],  int   * sa,  int   * sb ) {
 48           int  i, tmp;
 49          tmp  =   * sa;
 50           * sa  =   * sb;
 51           * sb  =  tmp;
 52           for  ( i  =   0 ; i  <  m;  ++ i ) {
 53                  tmp  =  a[ i ];
 54                  a[ i ]  =  b[ i ];
 55                  b[ i ]  =  tmp;
 56          }
 57  }
 58 
 59  void  solve() {
 60           int  sa, sb, sc, i;
 61           if  ( n  &   1  ) {
 62                  solveOdd();
 63          }
 64           else  {
 65                  solveEven();
 66          }
 67          sa  =  sb  =  sc  =   0 ;
 68           for  ( i  =   0 ; i  <  m;  ++ i ) {
 69                  sa  +=  a[ i ];
 70                  sb  +=  b[ i ];
 71                  sc  +=  c[ i ];
 72          }
 73           if  ( sa  <  sb ) {
 74                  swap( a, b,  & sa,  & sb );
 75          }
 76           if  ( sa  <  sc ) {
 77                  swap( a, c,  & sa,  & sc );
 78          }
 79           if  ( sb  <  sc ) {
 80                  swap( b, c,  & sb,  & sc );
 81          }
 82  }
 83 
 84  void  output() {
 85           int  i;
 86          printf(  " %d\n " ,  3 * n - 3  );
 87           for  ( i  =   0 ; i  <  m;  ++ i ) {
 88                  printf(  " %d %d\n " , c[ i ], b[ i ] );
 89          }
 90          printf(  " %d " ,  3 * n - 4  );
 91           for  ( i  =   0 ; i  <  m;  ++ i ) {
 92                  printf(  "  %d " , a[ i ] );
 93          }
 94          printf(  "  %d\n " ,  3 * n - 5  );
 95  }
 96 
 97  int  main() {
 98           int  td;
 99          scanf(  " %d " ,  & td );
100           while  ( td --   >   0  ) {
101                  scanf(  " %d " ,  & n );
102                  solve();
103                  output();
104          }
105           return   0 ;
106  }
107