连续N天后计算价格P的预期增长的程序

给定一个整数P ，在接下来的连续N天中，它们分别以50％的概率增加A或B ，任务是在N天后找到期望值。

例子：

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

方法：按照以下步骤解决问题：

1. 每天的期望增加值=(增加的可能性，A)* A +(增加价值的可能性，B)* B = (1/2)* A +(1/2)* B。
2. 因此，一天后的价值增加= (a + b)/ 2。
3. 因此， N天后的价值增加= N *(a + b)/ 2。
4. 因此， N天后增加的值= P + N *(a + b)/ 2。
5. 将增加的值打印为所需的答案。

下面是上述方法的实现：

C++

// C++ program for
// the above approach
#include 
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
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

输出：

3450

时间复杂度： O(1)
辅助空间： O(1)