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# 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

#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 −

## Output

Count of a after n seconds for given reproduction rate is: 68 Count of b after n seconds for given reproduction rate is: 0Count of c after n seconds for given reproduction rate is: 0

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