What if a = 2 and b = 3? The answer will be 1 but your code will give 0. Now fix it.

first = int(input())
second = int(input())

count = 0
for i in range(first, second + (second % 2 != 0), 2):
    count += 1

print(count)

Janus brokes fuses every ODD day. There is 1 odd day in interval [2..4]

#include <iostream>

int main()
{
    int a,b;
    int answer=0;
    std::cin>>a>>b;
    if(a%2==0)a++;
    for(int i=a;i<=b;i+=2)
        answer++;
    std::cout<<answer;
    return 0;
}

Try this:
4 4
The answer should be 0.

Edited by author 02.04.2015 00:59

но это не тупой брутфорс, просто в четные дни он не ломает предохранители. 100-ый день четный, поэтому ответ 50, а не 51

(B-(A/2*2+1)+2)/2

Seriously, guys, put away correct formulas, please.

If ((last day) mod 2 = 0) than formula = ((b - a)+1)/2
else formula = ((b - a) + 2) / 2
I got AC
Because if we our seconf day mod = 0, it means, that we mustn't count last day, only add 1. If our second day mod 2 = 1, it means, that we must count second day, add 2.
That's easy

Edited by author 14.10.2012 16:23

Edited by author 14.10.2012 16:24

It isn`t anything complex . It gives u the beginning and ending of an interval :
A and B
and wants you to calculate how many digits in this interval ( including the numbers A and B ) are odd. So u see it is just as simple. Good luck.

Edited by author 17.10.2004 21:12

The fusses blew up only in days where d%2==1 ( is something wrong with the problem description because it doesn't specify this ).

it's very easy to write after understand the problem. The problem is that one man is making something in every day after day.for example he will make something in first day then in third day then in 5th day an so on.you should calculate how many things he will do from A to B. A and B are integers to understand you can think that they are numbers of the day in year.
sorry for mu English.
Regards

You are wrong. Tests contain only two integers with exactly one end of line after each of them (no other characters).

speak English, respect other

I'd say this is a brute force. Here is mine AC:

var a,b,c:integer;
begin
  readln(a,b);
  c:=(b-a+1) div 2;
  if (odd(b)) and (odd(a)) then inc(c);
  writeln(c);
end.