C

(With apologies to Darren Stone)

long n,o,p,q,r;main(){while(--n){while(--o){while(--p){while(--q){while(--r){putchar('z'-'A');}}}}}}

n = 2^64 digit number (9...)

l = 100 chars of code

score = approximately 1e+2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936570

[ Score = n^5/l^3 = (10^(2^320)-1)/(100^3) = (10^2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576-1)/(10^6) ]

Note that I deserve to be flogged mercilessly for this answer, but couldn't resist. I don't recommend acting like me on stackexchange, for obvious reasons. :-P

EDIT: It would be even harder to resist the temptation to go with something like

long n;main(){putchar('z'-'A');putchar('e');putchar('+');while(--n){putchar('z'-'A');}

...but I suppose that an intended but unspecified rule was that the entire run of digits making up the number must be printed.

C

(With apologies to Darren Stone)

long n,o,p,q,r;main(){while(--n){while(--o){while(--p){while(--q){while(--r){putchar('z'-'A');}}}}}}

n = 2^64 digit number (9...)

l = 100 chars of code

score ≈ 1e+2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936570 ≈ 10↑↑3.2974890744

[ Score = n^5/l^3 = (10^(2^320)-1)/(100^3) = (10^2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576-1)/(10^6) ]

Note that I deserve to be flogged mercilessly for this answer, but couldn't resist. I don't recommend acting like me on stackexchange, for obvious reasons. :-P

EDIT: It would be even harder to resist the temptation to go with something like

long n;main(){putchar('z'-'A');putchar('e');putchar('+');while(--n){putchar('z'-'A');}

...but I suppose that an intended but unspecified rule was that the entire run of digits making up the number must be printed.

C

(With apologies to Darren Stone)

long n,o,p,q,r;main(){while(--n){while(--o){while(--p){while(--q){while(--r){putchar('z'-'A');}}}}}}

n = 2^64 digit number (9...)

l = 100 chars of code

score = approximately 1e+2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936570

[ Score = n^5/l^3 = (10^(2^320)-1)/(100^3) = (10^2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576-1)/(10^6) ]

Note that I deserve to be flogged mercilessly for this answer, but couldn't resist. I don't recommend acting like me on stackexchange, for obvious reasons. :-P

EDIT: It would be even harder to resist the temptation to go with something like

long n;main(){putchar('z'-'A');putchar('e');putchar('+');while(--n){putchar('z'-'A');}

...but I suppose that an intended but unspecified rule was that the number must actually be printed in full.

C

(With apologies to Darren Stone)

long n,o,p,q,r;main(){while(--n){while(--o){while(--p){while(--q){while(--r){putchar('z'-'A');}}}}}}

n = 2^64 digit number (9...)

l = 100 chars of code

score = approximately 1e+2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936570

[ Score = n^5/l^3 = (10^(2^320)-1)/(100^3) = (10^2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576-1)/(10^6) ]

Note that I deserve to be flogged mercilessly for this answer, but couldn't resist. I don't recommend acting like me on stackexchange, for obvious reasons. :-P

EDIT: It would be even harder to resist the temptation to go with something like

long n;main(){putchar('z'-'A');putchar('e');putchar('+');while(--n){putchar('z'-'A');}

...but I suppose that an intended but unspecified rule was that the entire run of digits making up the number must be printed.

C

(With apologies to Darren Stone)

long n,o,p,q,r;main(){while(--n){while(--o){while(--p){while(--q){while(--r){putchar('z'-'A');}}}}}}

n = 2^64 digit number (9...)

l = 100 chars of code

score = approximately 1e+2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936570

[ Score = n^5/l^3 = (10^(2^320)-1)/(100^3) = (10^2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576-1)/(10^6) ]

Note that I deserve to be flogged mercilessly for this answer, but couldn't resist. I don't recommend acting like me on stackexchange, for obvious reasons. :-P

C

(With apologies to Darren Stone)

long n,o,p,q,r;main(){while(--n){while(--o){while(--p){while(--q){while(--r){putchar('z'-'A');}}}}}}

n = 2^64 digit number (9...)

l = 100 chars of code

score = approximately 1e+2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936570

[ Score = n^5/l^3 = (10^(2^320)-1)/(100^3) = (10^2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576-1)/(10^6) ]

Note that I deserve to be flogged mercilessly for this answer, but couldn't resist. I don't recommend acting like me on stackexchange, for obvious reasons. :-P

EDIT: It would be even harder to resist the temptation to go with something like

long n;main(){putchar('z'-'A');putchar('e');putchar('+');while(--n){putchar('z'-'A');}

...but I suppose that an intended but unspecified rule was that the number must actually be printed in full.