#include<bits/stdc++.h>
#define ll long long int
using namespace std;
ll a[1001];
int main()
{
ll i, j, n, m, b;
i = 0;
vector<ll>a;
while(cin >> b)
{
a.push_back(b);
sort(a.begin(), a.end());
if(i%2 == 1)
{
n = i/2;
m = (a[n] + a[n+1])/2;
cout << m << endl;
}
if(i%2 == 0)
{
n = i/2;
cout << a[n] << endl;
}
i++;
}
return 0;
}
#define ll long long int
using namespace std;
ll a[1001];
int main()
{
ll i, j, n, m, b;
i = 0;
vector<ll>a;
while(cin >> b)
{
a.push_back(b);
sort(a.begin(), a.end());
if(i%2 == 1)
{
n = i/2;
m = (a[n] + a[n+1])/2;
cout << m << endl;
}
if(i%2 == 0)
{
n = i/2;
cout << a[n] << endl;
}
i++;
}
return 0;
}
pleease explain this solution
ReplyDeleteDear Sir,
ReplyDeletefor solve this problem first we will be know what is Median?
median of this array [1, 2, 3, 4, 5, 6, 7, 8, 9] is -------> 5. because array length is odd.
On the other hand.........
median of this array [1, 2, 3, 4, 5, 6, 7, 8] is ----------> (4+5)/2 = 4. because array length is even.
In this problem, first given 1. so median is 1.
second given 3. so the array is [1, 3] and median is -----> (1+3)/2 = 2. because array length is even
third given 4. so the array is [1, 3, 4] and median is ------> 3. because array length is odd
fourth given 60. so the array is [1, 3, 4, 60] and median is --------> (3+4)/2 = 3.
fifth given 70. so the array is [1, 3, 4, 60, 70] and median is --------> 4.
.................. so on.......
I hope you understand this problem........... Happy Coding....