Dr. Vishnu is opening a new world class hospital in a small town designed to be the first preference of the patients in the city. Hospital has N rooms of two types - with TV and without TV, with daily rates of R1 and R2 respectively. However, from his experience Dr. Vishnu knows that the number of patients is not constant throughout the year, instead it follows a pattern. The number of patients on any given day of the year is given by the following formula – (6-M)^2 + |D-15| where M is the number of month (1 for jan, 2 for feb ...12 for dec) and D is the date (1,2...31). All patients prefer without TV rooms as they are cheaper, but will opt for with TV rooms only if without TV rooms are not available. Hospital has a revenue target for the first year of operation. Given this target and the values of N, R1 and R2 you need to identify the number of TVs the hospital should buy so that it meets the revenue target. Assume the Hospital opens on 1st Jan and year is a non-leap year. Constraints Hospital opens on 1st Jan in an ordinary year 5 <= Number of rooms <= 100 500 <= Room Rates <= 5000 0 <= Target revenue < 90000000 Input Format First line provides an integer N that denotes the number of rooms in the hospital Second line provides two space-delimited integers that denote the rates of rooms with TV (R1) and without TV (R2) respectively Third line provides the revenue target Output Minimum number of TVs the hospital needs to buy to meet its revenue target. If it cannot achieve its target, print the total number of rooms in the hospital. Timeout 1 Test Case Example 1 Input 20 1500 1000 7000000 Output 14 Explanation Using the formula, number of patients on 1st Jan will be 39, on 2nd Jan will be 38 and so on. Considering there are only twenty rooms and rates of both type of rooms are 1500 and 1000 respectively, we will need 14 TV sets to get revenue of 7119500. With 13 TV sets Total revenue will be less than 7000000 Example 2 Input 10 1000 1500 10000000 Output 10 Explanation In the above example, the target will not be achieved, even by equipping all the rooms with TV. Hence, the answer is 10 i.e. total number of rooms in the hospital.
Answers
Answer:
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Answer:
Program is generated based on the given statement,
Step-by-step explanation:
Approach:
- The idea is to traverse through the whole year and generate revenue for each day. Find the revenue for the first year for every possible number of rooms that can have a TV. Follow the below steps to solve the problem:
- Suppose the total number of rooms having a TV is tvs, where 0 ≤ tvs ≤ N.
- For each number for TVs, revenue for the first year can be found by traversing through each day.
- The above formula can be used to find the number of patients per day. Suppose np is the current number of patients.
- The current number of patients np can have the value of the number of patients today or the total number of rooms whichever is minimum.
- If the number of patients is less than the number of rooms without TV, then the total revenue for today will be np*R2 because it is cheap.
- Else if the number of patients is greater than the rooms without TV, then the total revenue for today is given by:
- (N – tvs)*R2 + (np – (N – tvs))*R1
- Add the revenue for each day in target value and print the maximum target value generated after checking all possible combinations.
Program:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns number of // patient for a day in a month
int getPatients(int M, int D)
{
return ((6 - M) * (6 - M)) + abs(D - 15);
}
// Function that count the TVs with // given amount of revenue target
void countTVs(long long n, long long r1, long long r2, long long target)
{
long long np, tvs, current_target;
// Days in each month
vector<int> days = { 0, 31, 28, 31, 30, 31, 30, 31,31, 30, 31, 30, 31 };
// Check all possible combinations
for (tvs = 0; tvs <= n; tvs++) {
// Stores the current target
current_target = 0;
for (int m = 1; m <= 12; m++) {
for (int d = 1;
d <= days[m]; d++) {
// Number of patients // on day d of month m
np = getPatients(m, d);
// Patients cannot be // exceed number of rooms
np = min(np, n);
// If the number of patient is // <= count of rooms without tv if (np <= n - tvs) {
// All patients will opt // for rooms without tv
current_target += np * r2;
}
// Otherwise
else {
// Some will opt for // rooms with tv and
// others without tv
current_target += ((n - tvs) * r2 + ((np - (n - tvs))* r1));
}
}
}
// If current target meets // the required target
if (current_target >= target) {
break;
}
}
// Print the count of TVs
cout << min(tvs, n);
}
// Driver Code
int main()
{
long long N = 20, R1 = 1500;
long long R2 = 1000;
long long target = 7000000;
// Function Call
countTVs(N, R1, R2, target);
return 0;
}