Move 1988-2004 entries into subdirs
We we move 1988-2004 entries from all of a year's winning entries in a single directory to each winner being in their own sub-directory under the given year directory.
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Landon Curt Noll committed
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@@ -0,0 +1,184 @@
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#include <time.h>
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#include <stdio.h>
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#include <stdlib.h>
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#define __ "Y\n : ! ,.?>"
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#define D(f) ; } int f(int i) { return
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#define E(a,b,c,d,e) b##d,c,e,a
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#define e(a,b,c,d) a##b##c##d
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#define B *p++ - ' ' || !p++
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#define _(k) o(*#k - 64)
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#define G(g,l,k,o,e,d,n,c,i,b,a,f,j,m,p) \
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a#b#c#d#e#f#g#a#i#j#k#l#m#n#o e,i l##c,m##b; d g j n h(k 30:1); o p f}
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G(E(f ( 9) -s
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? -s + 'q'
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? f( 5 ) -s &&'d' -
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s?U( 3 ) ||_(J)
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:A ( s- f( 5 ))
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:(c= 4 *
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4):
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g(0)D(
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V)c <= 3 *4 + 3&&c && V(X ( i) ) ;
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}
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int , time ( 0)& , c - 15 || P
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(1
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):0,
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& ( !!d ( 0 )
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||(a=! _(S)) ,b= 0 )||
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a
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?d(a = 0)
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? _
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(I): w ( j =!W(j=_ (V ))) ,
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l(_
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(Y+ s ) )
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)e
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(
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main(){;
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;;;s , ra,
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n, d), && C ( j +R( s-j+t( i- 1) )
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)D( W)m[ c* 4| z( j ) ]
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&&_( X )E (d ( 0
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)&& R( 6)/ 6
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? c= 17|_ ( T )
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:01,||+
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, v =i & 3 ^ 2
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j ? k ( O[j]=R(15 ))
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i?
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c=i /4 & 63,
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a= 1
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D ( f )( p =M ,q ( i),*p+86)
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D ( l) * f, B?
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B ?B ?B? B ?B ?B
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?7 :B?B ?36: 59 * 89 *5 * 2 :B?6587
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*77 :9 + 2:B ?B?B?9: 5 : 37 : B ? 5 +5:4:20:
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B?B?B?32
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:e( 4, 3 , 1 *4 * 59 * 1 , 7):B?
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17: 34 : 19 :B ?
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8 *2 :B ?
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2:26: B?
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B?B ?B ?
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82 *+ 576: B ?33 :22 :5 * 5:3 *4 :3:B ?B ?B? B ?6 +
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+ 17 : B?B?13:6 : 9 *2:B?15 : 7 *2 :
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B ?
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B ?
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27:24: 31 : B?
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B? + 29 :,
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m [a^ 2 ] =T( 8 * 7) ;j= 3
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;
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;
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Q, ;int
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P ( int
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i )
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{time_t(x ) =+ 0+
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time(0) ; j=E( u (i ) : i D(R )e(r ,
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a ,n ,d) ()% i+1, ( e( l , o,
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caltim , e ) (& x)) ->e( tm_h , o, u , r)
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;retur , i?_( W
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): _( U) D(Y) *p&& *p<3 * 11 ?
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( ++ p ), n j= + 34 +( j< 6
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||j>19 ),Y (i)
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: *p
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D ( F) i == 9 ?1 :
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F( (i+ 7)% ( 7 + 3) ) +!O[ i
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]D(u) *p &&* p -32? p ++)
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D (h ) i ?*++*S=i% ( 38)- 1,
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D ( A )! n?i&&!g ( (0 ) ) ||K ( j=4| ! !i ) : N(!i )D(
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(s ) &&Q ( t( i)^i )
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D( U )i>= 0 ?U( i- 1 )||~f (i+1 ) == ~ s
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&&H (m[c* 4 |z(i)] )
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:0;D
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( X) + b && b<, for (*S = M ; *
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p; ),
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gets ( L, + 98 , stdin ) ?
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p= L, E (r( 0 ): 0 ,h( i / 38 ) : 0;D( C)!
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( m[j]+m [a= i] ) ?m[j[m]= i]=j:0;D (d) !
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b+(* O-
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c)*
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* O; D (z )
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-~- ~ -~i +
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v
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& 3 D(y)( putchar ) ( i)
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D ( q ) i& ,!(O[j = i]
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|| _ ( O))+a :0;
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D (w)O[ j] ==c && o
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(( !i ++ ? ! !_ (M
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):
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0) +15)|| ++ j<9 &&w (
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i)D ( r )*p? *p -4?
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* p -6 ?
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y(*p <9 + 2?__[
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* p
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]: * p+ 7 +79):o( j+1) : o (26+j),&q ( -!
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* p ++ + i )D(x)i? a=x ( ( -- i)), ++ p )
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D(J) !j --E( 0 D ( K )!-!N, || i &,
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_( R ) :i* c== (O
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[ j ] ) &&n==f (j+ 6 + 5 * 4)?O
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[j]= !i * c-_ ( L) :J(i )
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D ( N ) j=9, & d(0 )? j= 6 | ! i , J(i )D
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( o)(*S = p , ++ S,p = M , r (q (i) ),-- S, p = *S))
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( (l (_ (Q )),n =s,i ^5 ) ) ;, E(j ,
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sign , v ,ed ( b ) ;;
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int c =1 ,O[9])
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;int( s
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)=
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E (n ;char, 1 ,a,6 ,o ())* Z[
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E(M [37 * 9 *3 ] ,0,L [ 99] ,3] , * *S=Z ) ;
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E (*p=+1 ,char( m)[6 , co , 4 ],ol)
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+,
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V ( v=*O = k( j= 9 )) ;
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return ! c&& - ~ ~ - -! printf
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( ( "%d \n") , F (P ( 0) +6) *50 + ! ! b*
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( +b - (int)time ( 0 )) )
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;, ( b= 60 + time ( 0)+ 60 * 4) ;, n =Y
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(u (s =Y (i) ) )
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: (exit( 0)
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,i)D( g )x ( _(N) ^ 9)? !( a =0) : _ ( P )D (T)! i|,
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(2); s=35; Q(3);
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s=59; Q(6);
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a=1;
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)
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