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  1.  
  2.  
  3. i,j,m;f(d,o,n)int*d,*o;{for(i=n;i--;)for(m=d[!i],j=n;j;o[i]=m=--j^i&&d[j]<m?d[j]:m);}
  4.  
  5. int main( void )
  6. {
  7.     struct
  8.     {
  9.         int* integer ;
  10.         int count ;
  11.     }
  12.     d[] = {
  13.                 { ( int[] ){ 4 , 3 , 2 , 5 } , 4 } ,
  14.                 { ( int[] ){ 4 , 2 , 2 , 5 } , 4 } ,
  15.                 { ( int[] ){ 6 , 3 , 5 , 5 , 8 } , 5 } ,
  16.                 { ( int[] ){ 7 , 1 } , 2 } ,
  17.                 { ( int[] ){ 9 , 9 } , 2 } ,
  18.                 { ( int[] ){ 9 , 8 , 9 } , 3 } ,
  19.              } ;
  20.     const int c = sizeof( d ) / sizeof( d[0] ) ;
  21.  
  22.     for( int p = 0 ; p < c ; p++ )
  23.     {
  24.         int out[d[p].count] ;
  25.  
  26.         f( d[p].integer , out , d[p].count ) ;
  27.  
  28.         for( int i = 0 ; i < d[p].count ; i++ )
  29.         {
  30.             printf( "%d " , out[i] ) ;
  31.         }
  32.         puts( "" );
  33.     }
  34.  
  35. }
Success #stdin #stdout 0s 9432KB
comments (?)
stdin
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Standard input is empty
stdout
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2 2 3 2 
2 2 2 2 
3 5 3 3 3 
1 7 
9 9 
8 9 8
https://ideone.com/oEtVbb
language:
C (gcc 8.3)
created:
7 years ago
visibility:
public

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