Answer:
#include <iostream>
using namespace std;
int main() {
int k;
double d;
string s;
cin >> k >> d >> s;
cout << s << " " << d << " " << k << "\n" << k << " " << d << " " << s; }
Explanation:
k is int type variable that stores integer values.
d is double type variable that stores real number.
s is string type variable that stores word.
cin statement is used to take input from user. cin takes an integer, a real number and a word from user. The user first enters an integer value, then a real number and then a small word as input.
cout statement is used to display the output on the screen. cout displays the value of k, d and s which entered by user.
First the values of k, d and s are displayed in reverse order. This means the word is displayed first, then the real number and then the integer separated again by EXACTLY one space from each other. " " used to represent a single space.
Then next line \n is used to produce a new line.
So in the next line values of k, d and s are displayed in original order (the integer , the real, and the word), separated again by EXACTLY one space from each other.
The program along with the output is attached.
Answer:
Time Complexity of Problem - O(n)
Explanation:
When n= 1024 time taken is t. on a particular computer.
When computer is 8 times faster in same time t , n can be equal to 8192. It means on increasing processing speed input grows linearly.
When computer is 8 times slow then with same time t , n will be 128 which is (1/8)th time 1024.
It means with increase in processing speed by x factor time taken will decrease by (1/x) factor. Or input size can be increased by x times. This signifies that time taken by program grows linearly with input size n. Therefore time complexity of problem will be O(n).
If we double the speed of original machine then we can solve problems of size 2n in time t.
Answer:
Social media influencing tactics has impacted the way I see social media posts in your daily life in the sense that if influences what posts I see and what posts I interact with.
Explanation:
Some examples of social media influencing tactics include ads, videos, social media influencers and so on.
When for example, and ad pops up on my social network and I click on it, similar ads of different products would appear later on my social network. In this way, my interaction with that ad has brought up similar ads to my attention. Likewise interaction with videos and social media influencers. When i interact with one video or social media influencer, a similar video or social media influencer is brought up to my attention in my feed.
Answer:
Convergent Plate Boundary (subduction zone)
Explanation:
A convergent plate boundary is a location where two tectonic plates are moving toward each other, often causing one plate to slide below the other (in a process known as subduction). The collision of tectonic plates can result in earthquakes, volcanoes, the formation of mountains, and other geological events. An example is the one the formed Andes Mountain.
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Answer:
// here is code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n,no_open=0;
cout<<"enter the number of lockers:";
// read the number of lockers
cin>>n;
// initialize all lockers with 0, 0 for locked and 1 for open
int lock[n]={};
// toggle the locks
// in each pass toggle every ith lock
// if open close it and vice versa
for(int i=1;i<=n;i++)
{
for(int a=0;a<n;a++)
{
if((a+1)%i==0)
{
if(lock[a]==0)
lock[a]=1;
else if(lock[a]==1)
lock[a]=0;
}
}
}
cout<<"After last pass status of all locks:"<<endl;
// print the status of all locks
for(int x=0;x<n;x++)
{
if(lock[x]==0)
{
cout<<"lock "<<x+1<<" is close."<<endl;
}
else if(lock[x]==1)
{
cout<<"lock "<<x+1<<" is open."<<endl;
// count the open locks
no_open++;
}
}
// print the open locks
cout<<"total open locks are :"<<no_open<<endl;
return 0;
}
Explanation:
First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.
Output:
enter the number of lockers:9
After last pass status of all locks:
lock 1 is open.
lock 2 is close.
lock 3 is close.
lock 4 is open.
lock 5 is close.
lock 6 is close.
lock 7 is close.
lock 8 is close.
lock 9 is open.
total open locks are :3
