An oil factory has N number of containers and
each has a different capacity. During
renovation, the manager decided to make
some changes with the containers. He wishes
to make different pairs for the containers in
such a way that in the first pair the container
of maximum capacity is paired with the
container of minimum capacity, and so on for
the rest of the containers, to maintain a
balance throughout all the pairs of containers.
Write an algorithm to make different pairs of
containers in such a way that the first
container in the pair is of maximum capacity
and second container in the pair is of minimum
capacity. write the c program
Answers
Answered by
17
#include <stdio.h>
void main {
int N;
int max ;
int min ;
int a[] ;
for (int i = 0 ; i <= N ; i++){
a[i] = max ;
a[i+1]= min ;
}
printf (a) ;
}
Answered by
0
Answer:
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int arr[n];
for(int i=0; i<n; i++){
cin>>arr[i];
}
for(int i=0; i<n; i++){
for(int j = i; j<n-1; j++){
if(arr[j] < arr[j+1])
{
int temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}
}
for(int i=0; i<n; i++){
if(i != n-1){
cout<<arr[i]<<" "<<arr[n-1]<<endl;
n = n-1;
}
else{
cout<<arr[i]<<" "<<"0"<<endl;
}
}
return 0;
}
Explanation:
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