#include <iostream>

int main()
{
    int k1,k2;
    int count=0;
    int flag=1;


    std::cin>>k1>>k2;

    for(;;)
    {
        if(flag==1)
        {
            if(count==k1)
            {
                std::cout<<"yes";
                return 0;
            }
            count++;
            flag=-flag;
        }
        if(flag==-1)
        {
            if(count==k2)
            {
                std::cout<<"yes";
                return 0;
            }
            count++;
            flag=-flag;
        }
        if(count>k1 && count>k2)
        {
            std::cout<<"no";
            return 0;
        }
    }


}

a better way:
#include<iostream>
using namespace std;
int main(){
    int lock1,lock2;
    cin>>lock1>>lock2;
    if(lock1%2==0||lock2%2==1){
        cout<<"yes";
    }else{
        cout<<"no";
    }
    return 0;
}

What's the meaning of your code ? and I'm not sure what's the main method to steal the bike by the thief.
Could you mind tell me ?

Edited by author 05.11.2012 22:48

you can see that thief is testing even code on night 1, 3, 5...
and odd code on night 2, 4, 6...

So thief can steal the bike if the code is even on night 1, 3, 5...
and odd code on night 2, 4, 6...

Code may be even on night 1,3,5 only if first lock have even code.
Code may be odd on night 2,4,6 only if second lock have odd code.

So you have to check only if first lock have even code or second have odd code.

1st. code is correct, but does not meet the requirements of the problem ..
based on the conditions of the problem, such a check should be:
if((a>0 && a <= 9999 && a%2 == 0)||(b>=0 && b <=9999 && b%2 ==1))
        cout << "yes";
    else
    cout << "no";
0000 excluded from the search because the code has been entered., + indicated that a strictly four digit string to the test unchecked ..
verdict .. flawed testing

Edited by author 12.02.2013 17:11

#include <iostream>
using namespace std;
int main(){
    int l,b;
    cin>>l>>b;
    l%2==0 || b%2==1 ? cout<<"yes" : cout<<"no";
}

Edited by author 29.09.2013 13:22

Edited by author 29.09.2013 13:22