To get everything on one line with two inputs is not (easily) achievable, as far as I know. The closest you can get is: print 'I have', a=input() print 'apples and', p=input() print 'pears. '
Answer:
A.)
arr[0] = 10;
arr[1] = 10;
Explanation:
Given the array:
arr = {1,2,3,4,5}
To set the first two elements of array arr to 10.
Kindly note that ; index numbering if array elements starts from 0
First element of the array has an index of 0
2nd element of the array has an index of 1 and so on.
Array elements can be called one at a time using the array name followed by the index number of the array enclosed in square brackets.
arr[0] = 10 (this assigns a value of 10 to the index value, which replace 1
arr[1] = 10 (assigns a value of 10 to the 2nd value in arr, which replaces 2
Class B would be the correct answer.
Answer:
Explanation:
#include<iostream>
#include<ctime>
#include<bits/stdc++.h>
using namespace std;
double calculate(double arr[], int l)
{
double avg=0.0;
int x;
for(x=0;x<l;x++)
{
avg+=arr[x];
}
avg/=l;
return avg;
}
int biggest(int arr[], int n)
{
int x,idx,big=-1;
for(x=0;x<n;x++)
{
if(arr[x]>big)
{
big=arr[x];
idx=x;
}
}
return idx;
}
int main()
{
vector<pair<int,double> >result;
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
int choice;
cin>>choice;
while(choice!=2)
{
int n,m;
cout<<"Enter N"<<endl;
cin>>n;
cout<<"Enter M"<<endl;
cin>>m;
int c=m;
double running_time[c];
while(c>0)
{
int arr[n];
int x;
for(x=0;x<n;x++)
{
arr[x] = rand();
}
clock_t start = clock();
int pos = biggest(arr,n);
clock_t t_end = clock();
c--;
running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;
}
double avg_running_time = calculate(running_time,m);
result.push_back(make_pair(n,avg_running_time));
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
cin>>choice;
}
for(int x=0;x<result.size();x++)
{
cout<<result[x].first<<" "<<result[x].second<<endl;
}
}
