4 formatting

# C, 522

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }


## C 522

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }


# C, 522

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }

3 reformatted triangle

## C 522

A self demonstrating C answer. Couldn't be clearer! Bonus points for finding the extra character.

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }


## C 522

A self demonstrating C answer. Couldn't be clearer! Bonus points for finding the extra character.

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }


## C 522

A self demonstrating C answer. Couldn't be clearer! Bonus points for finding the extra character.

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }

2 rewrote intro

## C 522

Sadly, there is still one error in this code, but I would have to use another define to fix it, and I already claimed three. Since we could take args however we wanted, it takes the number of arguments as the number of lines to print outA self demonstrating C answer. BonusCouldn't be clearer! Bonus points for finding what line has too many charactersthe extra character.

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }


## C 522

Sadly, there is still one error in this code, but I would have to use another define to fix it, and I already claimed three. Since we could take args however we wanted, it takes the number of arguments as the number of lines to print out. Bonus points for finding what line has too many characters.

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }


## C 522

A self demonstrating C answer. Couldn't be clearer! Bonus points for finding the extra character.

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i][0]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }

1