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[Medium] C++ crack

By em, in Challenges (CTF)

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em Newbie

em

Posted
Posted (edited)

Se d? codul:


#include <iostream>
#include <cmath>
int main()
{
    char input[7]={0};
    std::cin>>input;
    input[0]=input[0]*input[2]-input[1]*input[3]-input[4]-input[4]%input[6];
    input[1]=(input[4]*input[5])%(--(--input[2]));
    input[2]=(input[5]+input[6])-(input[2]%100); 
    input[3]=((input[0]+input[1])-4)>>(input[2]/100);
    input[4]=((int) pow((++input[6]>>1),2)/3)-(input[6]+input[1]%76);
    input[5]=(input[5]+input[6])-input[2]/10;
    input[6]=((input[0]+input[1]*input[2])>>7)%((input[5]+2)>>1);

    std::cout<<input;
}

Care este inputul astfel incat output sa fie winner!

Limite input : exact 7 caractere ascii.

Face?i cum vre?i. Nu va uitati la rezolvarile altora si postati si voi acelasi lucru. :)

Solvers:

- tromfil

- CyberWolf08

Edited by em
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Zamolxis666 Newbie

Zamolxis666

Posted
Posted 


#include <iostream>
#include <cmath>
int main()
{
    char input[7]={0};
    std::cin>>input;
    /*input[0]=input[0]*input[2]-input[1]*input[3]-input[4]-input[4]%input[6];
    input[1]=(input[4]*input[5])%(--(--input[2]));
    input[2]=(input[5]+input[6])-(input[2]%100); 
    input[3]=((input[0]+input[1])-4)>>(input[2]/100);
    input[4]=((int) pow((++input[6]>>1),2)/3)-(input[6]+input[1]%76);
    input[5]=(input[5]+input[6])-input[2]/10;
    input[6]=((input[0]+input[1]*input[2])>>7)%((input[5]+2)>>1);*/

    strcpy(input, "winner");

    std::cout<<input;
}

Am facut cum am vrut :))

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em Newbie

em

Posted
  • Author
Posted

Eh. Prin cum vrei m-am referit ca o poti gandi fie matematic, fie faci un BK. (Pentru BK iti dau un indiciu ca parola mea este doar din litere mici + caracterele '-','_','O')

Oricum alegi sa o faci, ai ceva de invatat. (sper)

P.S.: Parola mea e parola mea, nu e unica desigur. :)

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TomescuMihail Newbie

TomescuMihail

Posted
Posted

:)) Inteligent, la asta chiar nu ma gandisem!


#include <iostream>
#include <cmath>
int main()
{
    char input[7]={0};
    std::cin>>input;
    /*input[0]=input[0]*input[2]-input[1]*input[3]-input[4]-input[4]%input[6];
    input[1]=(input[4]*input[5])%(--(--input[2]));
    input[2]=(input[5]+input[6])-(input[2]%100); 
    input[3]=((input[0]+input[1])-4)>>(input[2]/100);
    input[4]=((int) pow((++input[6]>>1),2)/3)-(input[6]+input[1]%76);
    input[5]=(input[5]+input[6])-input[2]/10;
    input[6]=((input[0]+input[1]*input[2])>>7)%((input[5]+2)>>1);*/

    strcpy(input, "winner");

    std::cout<<input;
}

Am facut cum am vrut :))

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em Newbie

em

Posted
  • Author
Posted
Aplicam metoda de la Defcamp: http://i40.tinypic.com/2464aco.png :-"

Nu mai face afi??rile. Merge mult mult mai repede dac? nu le mai faci. :P (Sau le faci doar cand se schimba prima liter?).

Again. Parola doar din litere mici + caracterele {-,_,O}. Ar fi OK dac? ar incerca si persoane care nu au mai implementat un BK in viata lor (sa posteze aici bucati de cod si sa fie ajutati).

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