Answer:
Information and communications technology is an extensional term for information technology that stresses the role of unified communications and the integration of telecommunications and computers, as Technology Trends 2016
#1: Spreading intelligence throughout the cloud. ...
#2: Self-managing devices. ...
#3: Communication beyond sight and sound. ...
#4: Fundamental technologies reshaping what networks can do. ...
#5: Weaving security and privacy into the IoT fabric.
Explanation:
look for a question that i have answered answer it and also plz give me brainliest on this one plz
Answer:
Correctly position the platen of the printer to hold the paper in place.
Explanation:
Since the dot matrix printer is printing dark and clear on the left of the paper but very light on the right side of the paper, it is possible that the paper is not being held properly. So adjusting the platen will hold the paper properly and hence uniform prints on the paper.
Note: Dot matrix printers are printers that print closely related dots to form require texts (shapes) by striking some pins against an ink ribbon.
Some of the parts of these printers are; power supply, carriage assembly, Paper sensor, ribbon, platen and pins.
Yes, mobile devices have to work in limited screen space.
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:
Each time you insert a new node, call the function to adjust the sum.
This method has to be called each time we insert new node to the tree since the sum at all the
parent nodes from the newly inserted node changes when we insert the node.
// toSumTree method will convert the tree into sum tree.
int toSumTree(struct node *node)
{
if(node == NULL)
return 0;
// Store the old value
int old_val = node->data;
// Recursively call for left and right subtrees and store the sum as new value of this node
node->data = toSumTree(node->left) + toSumTree(node->right);
// Return the sum of values of nodes in left and right subtrees and
// old_value of this node
return node->data + old_val;
}
This has the complexity of O(n).
Explanation:
