0 

#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;}
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2 Answers

0

It's confusing, hope this will help

#include <stdio.h>
int leech;
int main(int omg,char **Our,int Inferno) {
    char c,*D=Our[1];
    if(omg>0){
    	for(leech=0;D[leech];D[leech++]-=10){
    		D[leech++]-=120;
    		D[leech]-=110;
    		while (!main(0,Our,leech)) D[leech]+=20;
    		putchar((D[leech]+1032)/20);
    	}
    	putchar(10);
    }
    else{
    	c=omg+(D[Inferno]+82)%10-(Inferno>leech/2)*(D[Inferno-leech+Inferno]+72)/10-9;
    	D[Inferno]+=Inferno<0?0:!(omg=main(c/10,Our,Inferno-1))*((c+999)%10-(D[Inferno]+92)%10);
    	}
    return omg;
}
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-1

Không làm gì hết =))

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