Given an integer P, that increases either by A or B with 50% probability each, in the next N consecutive days, the task is to find the expected value after N days.
Examples:
Input: P = 1000, A = 5, B = 10, N = 10
Output: 1075
Explanation:
Expected increased value after N consecutive days is equal to:
P + N * (a + b) / 2 = 1000 + 10 × 7.5 = 1000 + 75 = 1075.Input: P = 2000, a = 10, b = 20, N = 30
Output: 2450
Approach: Follow the steps to solve the problem:
- Expected value of increase each day = (Probability of increase by A) * A + (Probability of value increase by B) * B = (1 / 2) * A + (1 / 2) * B.
- Therefore, increase in value after one day = (a + b) / 2.
- Therefore, increase in value after N days = N * (a + b) / 2.
- Therefore, increased value after N days = P + N * (a + b) / 2.
- Print the increased value as the required answer.
Below is the implementation of the above approach:
C++
// C++ program for // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the increased // value of P after N days void expectedValue( int P, int a,
int b, int N)
{ // Expected value of the
// number P after N days
double expValue
= P + (N * 0.5 * (a + b));
// Print the expected value
cout << expValue;
} // Driver Code int main()
{ int P = 3000, a = 20, b = 10, N = 30;
expectedValue(P, a, b, N);
return 0;
} |
Java
// Java program for the above approach import java.io.*;
class GFG{
// Function to find the increased // value of P after N days static void expectedValue( int P, int a,
int b, int N)
{ // Expected value of the
// number P after N days
double expValue = P + (N * 0.5 * (a + b));
// Print the expected value
System.out.print(expValue);
} // Driver code public static void main(String[] args)
{ int P = 3000 , a = 20 , b = 10 , N = 30 ;
expectedValue(P, a, b, N);
} } // This code is contributed by abhinavjain194 |
Python3
# Python3 program for # the above approach # Function to find the increased # value of P after N days def expectedValue(P, a, b, N):
# Expected value of the
# number P after N days
expValue = P + (N * 0.5 * (a + b))
# Print the expected value
print ( int (expValue))
# Driver Code if __name__ = = '__main__' :
P = 3000
a = 20
b = 10
N = 30
expectedValue(P, a, b, N)
# This code is contributed by ipg2016107 |
C#
// C# program for the above approach using System;
class GFG{
// Function to find the increased // value of P after N days static void expectedValue( int P, int a,
int b, int N)
{ // Expected value of the
// number P after N days
double expValue = P + (N * 0.5 * (a + b));
// Print the expected value
Console.Write(expValue);
} // Driver code static void Main()
{ int P = 3000, a = 20, b = 10, N = 30;
expectedValue(P, a, b, N);
} } // This code is contributed by abhinavjain194 |
Javascript
<script> // Javascript program for // the above approach // Function to find the increased // value of P after N days function expectedValue(P, a, b, N)
{
// Expected value of the
// number P after N day
var expValue = P + (N * 0.5 * (a + b)) ;
return expValue;
}
// Driver code
var P = 3000
var a = 20
var b = 10
var N = 30
document.write(expectedValue(P, a, b, N));
</script> |
Output:
3450
Time Complexity: O(1)
Auxiliary Space: O(1)