Problem:- Write a Hackerrank Solution For Day 26: Nested Logic or Hacker Rank Solution Program In C++ For " Day 26: Nested Logic " or Hackerrank 30 days of code Java Solution:Day 26: Nested Logic solution or Hackerrank solution for 30 Days of Code Challenges or Hackerrank 30 days of code Java Solution, Day 26: Nested Logic solution, or C/C++ Logic & Problem Solving: Day 26: Nested Logic or Hacker Rank Solution For Day 26: Nested Logic.
Check This:- Hacker rank solution for Strings, Classes, STL, Inheritance in C++.
Explanation:- We have to solve the problem according to given task, the task is given below. There are 4 cases given in the task our solution should pass the all 4 cases according to condition and print the Hackos or fine. Let's take an example and try to find out the solution of this problem.
Task:- Nested Logic Full filled the given task.
1. If the book is returned on or before the expected return date, no fine will be charged (i.e.: fine = 0).
2. If the book is returned after the expected return day but still within the same calendar month and year as the expected return date, fine = 15 Hackos * (The number of days late).
3. If the book is returned after the expected return month but still within the same calendar year as the expected return date, the fine = 500 Hackos * (The number of days late).
4. If the book is returned after the calendar year in which it was expected, there is a fixed fine of 10000 Hackos
Example:- Suppose we have given below input. Input formate is first DD: MM: YYYY
Actual =====>15 12 2015
Expected ===>15 6 2015
Now according for our logic in the program.
diff = 15 - 15 = 0.
mdiff = 12 - 6 = 6.
ydiff = 2015 - 2015 = 0.
Now first condition is failed actual year - expected year so according to ternary operator second part will be executed, so according to second part (Month-Month1>0&&ydiff==0) this part is passed cause 12 - 6 = 6 and year difference(actual year - expected ) is 0 so Month difference ((actual year - expected) * 500). Month difference is 6 * 500 = 3000. So this is a Hackos we have to pay.
Also Check:- Geeksforgeeks solution for School, Basic, Easy, Medium, Hard in C++.
Submit Your Solution Here:- Click Here
Solution:-
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int Day, Month, Year, Day1, Month1, Year1, mdiff, ddiff, ydiff, diff;
cin>>Day>>Month>>Year>>Day1>>Month1>>Year1;
mdiff = Month - Month1;
ddiff = Day - Day1;
ydiff = Year - Year1;
diff=(Year-Year1>0)?10000:(Month-Month1>0&&ydiff==0)?mdiff*500:(Day-Day1>0 && ydiff==0)?ddiff*15:0;
cout<<diff<<endl;
return 0;
}
Output:-
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Check This:- Hacker rank solution for Strings, Classes, STL, Inheritance in C++.
Explanation:- We have to solve the problem according to given task, the task is given below. There are 4 cases given in the task our solution should pass the all 4 cases according to condition and print the Hackos or fine. Let's take an example and try to find out the solution of this problem.
Task:- Nested Logic Full filled the given task.
1. If the book is returned on or before the expected return date, no fine will be charged (i.e.: fine = 0).
2. If the book is returned after the expected return day but still within the same calendar month and year as the expected return date, fine = 15 Hackos * (The number of days late).
3. If the book is returned after the expected return month but still within the same calendar year as the expected return date, the fine = 500 Hackos * (The number of days late).
4. If the book is returned after the calendar year in which it was expected, there is a fixed fine of 10000 Hackos
Example:- Suppose we have given below input. Input formate is first DD: MM: YYYY
Actual =====>15 12 2015
Expected ===>15 6 2015
Now according for our logic in the program.
diff = 15 - 15 = 0.
mdiff = 12 - 6 = 6.
ydiff = 2015 - 2015 = 0.
Now first condition is failed actual year - expected year so according to ternary operator second part will be executed, so according to second part (Month-Month1>0&&ydiff==0) this part is passed cause 12 - 6 = 6 and year difference(actual year - expected ) is 0 so Month difference ((actual year - expected) * 500). Month difference is 6 * 500 = 3000. So this is a Hackos we have to pay.
Also Check:- Geeksforgeeks solution for School, Basic, Easy, Medium, Hard in C++.
Submit Your Solution Here:- Click Here
Solution:-
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int Day, Month, Year, Day1, Month1, Year1, mdiff, ddiff, ydiff, diff;
cin>>Day>>Month>>Year>>Day1>>Month1>>Year1;
mdiff = Month - Month1;
ddiff = Day - Day1;
ydiff = Year - Year1;
diff=(Year-Year1>0)?10000:(Month-Month1>0&&ydiff==0)?mdiff*500:(Day-Day1>0 && ydiff==0)?ddiff*15:0;
cout<<diff<<endl;
return 0;
}
Output:-
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