#include <iostream>

using namespace std;

int main()
{
    long long a = 0;
    long long b = a * (a + 1);
    while (1) {
        if (b == 7943722218936282)
            break;
        a++;
        b = a * (a + 1);
    }
    cout << a << endl;
    system("PAUSE");
    return 0;
}

public void onClick(View paramAnonymousView)
      {
        String str = localEditText.getText().toString();
        paramAnonymousView = paramBundle.getText().toString();
        if (str.equals(paramAnonymousView))
        {
          MainActivity.showmsg("user is equal passwd");
        }
        else if ((str.equals("admin") & paramAnonymousView.equals("pass71487")))
        {
          MainActivity.showmsg("success");
          MainActivity.showmsg("flag is flag{borring_things}");
        }
        else
        {
          MainActivity.showmsg("wrong");
        }
      }

First remove the confusion in the code and adjust the stack balance.

wrong function, the odd and even subscripts of flag are XOR subscript and minus subscript respectively.

omg function, transformed flag and unk_4030C0 comparison.

Reverse solution, flag{fak3_alw35_sp_me!!}

Use dbg to debug to

Here, flag{fak3_alw35_sp_me!!} and hahaha_ Do_ you_ find_ Exclusive or of the first 19 characters of me

[0x0E,0x0D,0x09,0x06,0x13,0x05,0x58,0x56,0x3E,0x06,0x0C,0x3C,0x1F,0x57,0x14,0x6B,0x57,0x59,0x0D,0x00]

Inverse solution

flag{d07abccf8a410c, 5 characters are missing, the last bit is'} '

In the finally function, the five digit values are used

It can be seen that 0x3a must be '}', the relationship between guesses is XOR (71), and a complete flag is obtained.

flag{d07abccf8a410cb37a}

You can't blow up the last few bits of this question, because you can't go through the check flag when you bring it in, and finally guess it's XOR with a bit of brain hole.

 

signal