3 ms·
Reminds me of 2001/cheong [1]. #include <stdio.h> int l;int main(int o,char **O, int I){char c,*D=O[1];if(o>0){ for(l=0;D[l ];D[l
by lifthrasiir 5y ago
Reminds me of 2001/cheong [1].
#include <stdio.h>
int l;int main(int o,char **O,
int I){char c,*D=O[1];if(o>0){
for(l=0;D[l ];D[l
++]-=10){D [l++]-=120;D[l]-=
110;while (!main(0,O,l))D[l]
+= 20; putchar((D[l]+1032)
/20 ) ;}putchar(10);}else{
c=o+ (D[I]+82)%10-(I>l/2)*
(D[I-l+I]+72)/10-9;D[I]+=I<0?0
:!(o=main(c/10,O,I-1))*((c+999
)%10-(D[I]+92)%10);}return o;}
[1] https://www.ioccc.org/2001/cheong.hint https://www.ioccc.org/2001/cheong.hint
- ducktective 5y ago`curl -s https://www.ioccc.org/2001/cheong.hint https://www.ioccc.org/2001/cheong.hint | nc termbin.com 9999` https://termbin.com/5yaq https://termbin.com/5yaq In short: for a 2n-digit input, returns the integer part of its square root (n-digits)
- Trung0246 5y agoIs there a tool that automatically generate minified code like this but with ascii art? Or do these code snippets are done by hands?
- lifthrasiir 5y agoThere are some tools, but in general it is easier to do so manually---you can alter the code when the picture doesn't seem to fit in the template, while tools generally can't.