Program to calculate expected increase in price P after N consecutive days
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)
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