there is a question to find middle number . i tried really hard [closed] - c++

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I need to find middle number from 3 numbers the user inputs. I'm completely new to C++.
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a, b, c, d, mdq, mnq, mxq, mdw, mnw, mxw, mde, mne, mxe;
cin>>a>>b>>c;
mda = c;
if (mda>a and mda<b) mdq = c;
if (mda>a and mda>b) mxa = c;
if (mdq<a and mdq<b) mnq = c;
mdw = a:
if (mdw>b and mdw<c) mdq = a;
if (mdw>b and mdw>c) mxg = a;
if (mdw<b and mdw<c) mnq = a;
mde = b;
if (mde>a and mde<c) mdq = b;
if (mde>a and mde>c) mxq = b;
if (mde<a and mde<c) mnq = b;
d = mdq+mxq+mng:
d = d-mxq-mnq;
cout<<d;
}
I tried to find middle number and it sometimes work but whenever i type 4 7 1 the output is 1.

Answer 1

I think you are trying far too hard. There are only six possibilities, you don't need complicated logic and you definitely don't need mathematics.
#include <iostream>
int main()
{
int a, b, c;
std::cin >> a >> b >> c;
// it's a if b <= a <= c OR c <= a <= b
if ((b <= a && a <= c) || (c <= a && a <= b))
std::cout << "first # " << a << '\n';
// it's b if a <= b <= c OR c <= b <= a
else if ((a <= b && b <= c) || (c <= b && b <= a))
std::cout << "second # " << b << '\n';
// neither of the above, must be c
else
std::cout << "third # " << c << '\n';
}
I suppose the only tricky part is how to deal with equal numbers. I used <= which means given equal numbers the earlier number will be printed. So for example given 11 22 22 the output is second # 22 not third # 22.

Answer 2

Consider:
#include <algorithm>
#include <iostream>
int main()
{
int a, b, c ;
std::cin >> a >> b >> c ;
if( a > b ) std::swap( a, b ) ;
if( b > c ) std::swap( b, c ) ;
if( a > b ) std::swap( a, b ) ;
std::cout << b << '\n' ;
return 0;
}
It sorts the three values in order so that the answer is always b.
An alternative that does not modify the input:
#include <algorithm>
#include <iostream>
int main()
{
int a, b, c ;
std::cin >> a >> b >> c ;
int max1 = std::max( a, b ) ;
int max2 = std::max( b, c ) ;
int max3 = std::max( a, c ) ;
int mid = std::min( max1, max2 ) ;
mid = std::min( mid, max3 ) ;
std::cout << mid << '\n' ;
return 0;
}
which uses just four additional variables rather than ten in your attempt. It works by determining the maximum of all possible pairs: a,b, b,c and a,c, giving three values that contain either the largest or second largest value - the second largest value of three is also the middle value of course, so the smaller of these three maxima is the middle value.
You need not rely on the min()/max() functions here - they can be replaced with for example the expressions a < b ? a : b ; and a > b ? a : b or even if( a > b ) max1 = a; else max1 = b; etc.
Perhaps a more obvious solution, given that there are only 6 possible orderings of a b c, and only three possible outcomes for the middle value, then you can explicitly test for each possibility. You only need to explicitly test for two possible outcomes, since if neither of those are true, the answer is the third possibility:
#include <iostream>
int main()
{
int a, b, c;
std::cin >> a >> b >> c ;
int mid = a ; // Initially assume a is the middle
// (i.e. order: b a c or c a b)
// If b is in the middle...
if( (a <= b && b <= c) || // a b c or...
(c <= b && b <= a) ) // c b a
{
mid = b ;
}
// else if c is in the middle
else if( (a <= c && c <= b) || // a c b or...
(b <= c && c <= a) ) // b c a
{
mid = c ;
}
std::cout << mid << '\n' ;
return 0 ;
}
This is essentially #john's solution (credit), but you may find the explanatory comments here easier to follow perhaps.