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;
}
}
Answer:
In Python:
def gcd(m,n):
if n == 0:
return m
elif m == 0:
return n
else:
return gcd(n,m%n)
Explanation:
This defines the function
def gcd(m,n):
If n is 0, return m
<em> if n == 0:
</em>
<em> return m
</em>
If m is 0, return n
<em> elif m == 0:
</em>
<em> return n
</em>
If otherwise, calculate the gcd recursively
<em> else:
</em>
<em> return gcd(n,m%n)</em>
<em />
<em>To call the function to calculate the gcd of say 15 and 5 from main, use:</em>
<em>gcd(15,5)</em>
Answer:
give leave taking clues
Explanation:
The feed forward part of the conversation should do all of the following except <u>give leave taking clues</u> a identify the tone introduce the purpose give leave-taking clues establish a time-frame
Answer:
April 25, 2020
Explanation:
April 25, 2020
In order for an investment to be considered long-term, it must be held for longer than 1 year.
April 24th is not longer than a year, the only answer that is longer than a year of holding is April 25, 2020
