Count of a, b & c after n seconds for given reproduction rate in C++

Given three numbers 'a', 'b' and 'c' as input. The goal is to find the count/value of 'a', 'b' and 'c' after n seconds such that the rate of reproductions are:-

Every a changes to b after every 2 seconds
Every b changes to c after every 5 seconds
Every c changes to 2 a after every 12 seconds.

Let us understand with examples.

For Example

Input - n_seconds = 62 a = 1 b = 1 c = 1

Output - Count of a after n seconds for given reproduction rate is: 0

Count of b after n seconds for given reproduction rate is: 33

Count of c after n seconds for given reproduction rate is: 1

Explanation - After 60 seconds, a's will be 32, b = 0, c = 0.

After 2 more seconds, all b's will become c, c=1. All a become b, b=33.

Input - n_seconds = 20 a = 1 b = 1 c = 1

Output - Count of a after n seconds for given reproduction rate is: 0

Count of b after n seconds for given reproduction rate is: 0

Count of c after n seconds for given reproduction rate is: 6

Explanation

1 sec :- a=1, b=1, c=1

2 sec :- a=0, b=2(1+1) , c=1 → a to b after 2 seconds

4 sec :- a=0, b=2 , c=1 → a to b after 2 seconds

5 sec :- a=0, b=0 , c=3 (1+2) → b to c after 5 seconds

6 sec :- a=0, b=0 , c=3 → a to b after 2 seconds

8 sec :- a=0, b=0 , c=3 → a to b after 2 seconds

10 sec :- a=0, b=0 , c=3 → b to c after 5 seconds

12 sec :- a=6 (0+2*3), b=0 , c=0 → c to 2a after 12 seconds

14 sec :- a=0, b=6(0+6) , c=0 → a to b after 2 seconds

15 sec :- a=0, b=0 , c=6(0+6) → b to c after 5 seconds

16 sec :- a=0, b=0 , c=6 → a to b after 2 seconds

18 sec :- a=0, b=0 , c=6 → a to b after 2 seconds

20 sec :- a=0, b=0 , c=6 → b to c after 5 seconds

Approach used in the below program is as follows

The LCM of seconds will be 60 ( 2, 5, 12 ). So the changes in a, b and c after every 60 seconds will be :

60 seconds → a= 32, b=0, c=0

120 seconds → a= 32*32 , b=0, c=0

180 seconds → a= 32*32*32 , b=0, c=0.

For seconds as multiples of 60 calculate a's as above. For non-multiples, calculate nearest multiple temp and then traverse from temp+1 to input seconds and calculate using mod 2, 5 or 12.

Take numbers a, b and c as input.
Take n_seconds as time in seconds.
Function reproduction_rate(int n_seconds, int a, int b, int c) takes all parameters and prints the count of a, b & c after n seconds for given reproduction rate.
Take temp = n_seconds / 60 as multiples of 60 below n_seonds.
Calculate a = (int)pow(32, temp) as given in formula. ( a^temp)
Now update temp = 60 * temp for nearest multiple of 60 less or equal to n_seconds.
Now traverse using for loop from i=temp+1 to i=n_seconds.
If the number i is multiple of 2 then update b by adding a and a with 0.
If the number i is multiple of 5 then update c by adding b and b with 0.
If the number i is multiple of 12 then update a by adding 2c and c with 0.
At the end of for loop, print the final values of a, b and c.

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;

void reproduction_rate(int n_seconds, int a, int b, int c) {
   int temp = n_seconds / 60;
   a = (int) pow(32, temp);
   temp = 60 * temp;

   for (int i = temp + 1; i <= n_seconds; i++) {
      if (i % 2 == 0) {
         b = b + a;
         a = 0;
      }
      if (i % 5 == 0) {
         c = c + b;
         b = 0;
      }
      if (i % 12 == 0) {
         a = a + (2 * c);
         c = 0;
      }
   }
   cout << "Count of a after n seconds for given reproduction rate is: " << a << "\n";
   cout << "Count of b after n seconds for given reproduction rate is: " << b << "\n";
   cout << "Count of c after n seconds for given reproduction rate is: " << c;
}
int main() {
   int n_seconds = 72;
   int a = 2;
   int b = 1;
   int c = 1;
   reproduction_rate(n_seconds, a, b, c);
   return 0;
}

If we run the above code it will generate the following output −