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miskamm [114]
2 years ago
15

Write a program that estimates how many years, months, weeks, days, and hours have gone by since Jan 1 1970 by calculations with

the number of seconds that have passed.
Computers and Technology
1 answer:
Rzqust [24]2 years ago
6 0

Answer:

Explanation:

The following code is written in Java. It uses the LocalDate import to get the number of seconds since the epoch (Jan 1, 1970) and then uses the ChronoUnit import class to transform those seconds into years, months, weeks days, and hours. Finally, printing out each value separately to the console. The output of the code can be seen in the attached picture below.

import java.time.LocalDate;

import java.time.temporal.ChronoUnit;

class Brainly {

   public static void main(String[] args) {

       LocalDate now = LocalDate.now();

       LocalDate epoch = LocalDate.ofEpochDay(0);

       System.out.println("Time since Jan 1 1970");

       System.out.println("Years: " + ChronoUnit.YEARS.between(epoch, now));

       System.out.println("Months: " + ChronoUnit.MONTHS.between(epoch, now));

       System.out.println("Weeks: " + ChronoUnit.WEEKS.between(epoch, now));

       System.out.println("Days: " + ChronoUnit.DAYS.between(epoch, now));

       System.out.println("Hours: " + (ChronoUnit.DAYS.between(epoch, now) * 24));    

   }

}

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Write an algorithm that prints out all the subsets of 3 elements of a set of n elements. The elements of the set are stored in a
Ksenya-84 [330]

void Print_3_Element_Subsets(int n,const element_type S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

print("No subset found");

}

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

for(int k =j+1; k < n ; k++)

{

print(S[i],S[j],S[k]);

}

}

}

}

C++ Implementation;

#include<cstdlib>

#include<time.h>

using namespace std;

int S1[5]= {1,2,3,4,5};

int n1 = 5;

int S2[2]={1,2};

int n2 = 2;

void Print_3_Element_Subsets(int n,const int S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

printf("No subset found\n");

}

printf("3 Subsets are\n");

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

for(int k =j+1; k < n ; k++)

{

cout<<S[i]<<" "<<S[j]<<" "<<S[k]<<endl;

}

}

}

}

int main()

{

cout<<"Case 1"<<endl;

Print_3_Element_Subsets(n1,S1);

cout<<endl<<"Case 2"<<endl;

Print_3_Element_Subsets(n2,S2);

}

OUTPUT

Case 1

3 Subsets are

1 2 3

1 2 4

1 2 5

1 3 4

1 3 5

1 4 5

2 3 4

2 3 5

2 4 5

3 4 5

Case 2

No subset found

3 Subsets are

--------------------------------

Process exited after 0.0137 seconds with return value 0

Press any key to continue . . .

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Answer:

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