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Lubov Fominskaja [6]
3 years ago
12

What will you see on the next line? >>> int(6.5)

Computers and Technology
2 answers:
tankabanditka [31]3 years ago
8 0

Answer:

6

Explanation:

The int functions founds down to the nearest whole number, which in this case would be 6.

Degger [83]3 years ago
6 0

Answer:

6

Explanation:

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Write a program to implement problem statement below; provide the menu for input N and number of experiment M to calculate avera
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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;

}

}

8 0
3 years ago
4) Which is a more efficient way to determine the optimal number of multiplications in a matrix-chain multiplication problem: en
Assoli18 [71]

Answer:

Running RECURSIVE-MATRIX-CHAIN is asymptotically more efficient than enumerating all the ways of parenthesizing the product and computing the number of multiplications of each.

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