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[Easy] C++ trick

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#include <iostream>
using namespace std;
int main()
{
//TODO
cout<<1[a]*2[b]-3[c];
return 0;
}

Completa?i codul de mai sus astfel încât programul s? afi?eze 31337.

Explica?i în câteva cuvinte de ce merge ce a?i f?cut voi.

Trimite?i solu?iile prin PM. S? zicem c? o s? câ?tige cel cu cea mai scurt? surs? (in bytes).

Nu exist? premii.

Cine a rezolvat pana acum?

- cifratorul (+ 32 bytes)

- Nytro (+ 34 bytes)

- razer_krait (+ 41 bytes)

- curtuy (+ 52 bytes)

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Câ?tig?torul este cifratorul. (+32bytes)


#include <iostream>
using namespace std;
int main()
{
int c[]={0,31337,1,0},*a=c,*b=c;
cout<<1[a]*2[b]-3[c];
return 0;
}

Explica?ie: In C++ a = *(a+B) = *(b+a) = b[a];

O solutie de +31 bytes ar fi : int c[]={0,3482,9,1},*a=c,*b=c;

Felicit?ri tuturor participan?ilor.

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