Landon Curt Noll
2ba12ecc3c
Under each yyyy/dir there is a file of the form '*.orig.c' containing the original source code, or as close as we can get to the original source code.
98 lines
3.9 KiB
C
98 lines
3.9 KiB
C
#include<stdio.h>
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#define c(C) printf("%c",C)
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#define C(c) ((int*)(C[1]+6))[c]
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main(int c,char
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*C[]) {(C[c]=C[
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1]+2 )[0]= c(52*c(\
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'C'+ '4'/4) );for(c
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=0; c<491;++ c)for(*
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*C= C[1]['c' +c] =
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0;* C[0]<8;( ** C
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)++ )C[1][c+ 'c']=
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*(C[ 1]+c+'c')+ C[1][
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99+ c]+(C[1 ][**C
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+8*c +99]==32 ); (
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*C)[4]=*C[2]== 75 ?
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*((C[2]+=3)-2 )==70?
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1:0:0;C(0)=C( 1)=c=0
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;while(*C[2]? C[2][1]
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?*(C[2]+2)?1 :0:0:0)
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{if( *C [2 ]>'w'){
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C(1)=0;C[1] [2]++;*C
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[2]=0;}else C(1)+=*C[
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2]==58?2+( C[2][3]&&
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*(C[2]+3)< 'x'):*C[2]
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=='s'?(C[ 2][1]-=48):
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*C[2]>=65 ?3-(*C[2]==\
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'm'?1:0) :1;C(0)=C(1)>
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C(0)?C(1 ):C(0);c+=3;*
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(C+2)+=3;}printf(" %d\
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%d\n", 56+8*C( 0),80**(C[3] ++))
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;*C[2]=0 ;C[2] -=c;*C[3] =0;
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while(C[3] [1,- 1]--){; for( **
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C=0 ;* *C< 80;(** C)++) {C
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[2] -=3 ** C[3]; *C[3] ++
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=0; *C[ 3] =**C>= 51||* *C<
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18 ||* *C %8!=2?0 :255 ;c(1
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-1 );c (*C [3]);for( (*C)[
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1] =0;( *C)[ 1]<3;(*C)[1] ++)c(*C
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[3 ]|(( *C)[ 4]?**C>18&&* *C<42 ?C[1][
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42 +*(* C+1) +3***C]:0: **C>= 11&&*
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*C <64? ~C[1 ][ 7***C+97 +(*C)[ 1]]:
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0) );c( *C[3 ]++) ;for(C (1)=0; (C(
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2) =C(1 ))<C (0);) {(*C) [2]=C [2][
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1] -49; c=(* C[2]<= 63); c=(* C)[0]
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-4 *(C[ 3][0 ]=105- C[2][ c] -7*(*(C
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[2]+c)< 'c') -18*( C[2][c ]<77)+2*(
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*C)[4 ]-7* (C[2] [c]<'C' ))-6;for(C(
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3)=0;
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C(3)< (*C[2]?*C[2
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]>'r' ?C[ 2][1]:(1
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+2*(*C [2]> 64) +(2-!C[2 ][3])
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*(58 == (*(C +2))[0])-
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(C[2] [0 ]=='m' ) ):C(2)?C(0 )-C(
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2):0 );C(3)++) C(1)+=c(C [1][4]|(* C[2]&&
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*C[2 ]<'s'?*C[2] ==58?C(3) ==1?**C >17&&**C
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<51 ?C[ 2] [1] ==59?39: C[2][1 ]==58?9:1:
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0:0 :63 >* C[2 ]?(c<7&& c>-9?C[1 ][(*C[2]<
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45? 'c' *5 +2* '%':*C[ 2]< 61?570 :571)+
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3*c ]: 0) :*C [2]>'o' ?**C>26&&**C<29 &&!
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(*C )[ 2] ||( *C)[2] ==1&&(&*C[0]) [0]
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<34 && 31 <** C?C(3) <2?15+225*C(3 ):0:(
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*C)[ 2] == 3?C (3)<2 &&**C>22&&**C <45&&C
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(3)< 2? C[ 1][7* *(*C)+151+C (3)]:0:
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7==( *C )[2 ]&&* *C>26&&**C <42&&C(3
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)<2?C [1 ][7 ***C +135+C(3)] :0:*C[2]
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<'k'? (c >-5 && c<5?C(3
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)<2?C [1][( (*C) [2]<3?
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207:205)+7* c+C(3 )]:C[
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2][2]==46 && (c==-2|| c==-
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1- 2*(*C[3]%2 ))?
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96 :0:0)|((*C)[2
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]?
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C( 3 ) ?C(3)< 2
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&& *C[3]> 7 &&
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c< 1&&c >-24 ?8:0:* C
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[3 ]<8&&c >1 &&c<24 ? ' '
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:0 :0) |(C(3) <2 &&( ** C
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==66 && *C[3 ]>14 ||* C[ 3
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]>12 && 58 ==**C|| *
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C[3 ]< 2 &&**C==10 ) ?
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5* 51: 0)|(7== (
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*C)[ 2]?*C[3] <
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8?c>13&& c<23&&C(3)<2?C
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[1][144+ 7*c+C(3)]:0:C(
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3) && c< -14&&c>- 24?C[
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1] [7 *c +400+C(3 )]:0:
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0) :! C (3)?**C> 21&&C
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[0 ][ 0] <32
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?C [ 1 ] [(
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* C ) [2 ] +323
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+7 ** *C ]:36
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+ 1 < ** C&& '0'>
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* *C ? C[1][
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C[ 2] [2]+162
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+ 7* **C]:0:0:0));C[1
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][ 3] ++;C[2]+=3;}c(0)
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;C [3 ]-=2;}*C[3]=0;}}
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