连续变化带来的总体变化百分比

给定一个表示百分比变化的数组Arr。任务是确定这些百分比更改后的百分比增加。

例子：

Input:  arr[] = {10, 20, 30, 10}
Output: Percentage change is = 88.76 %

Input:  arr[] = {20, 15, 9, 7}
Output: Percentage change is = 60.94 %

无需连续更改的解决方案：-

let us take a number N = 120.
and percentage changes are given as, arr[] = {10, 20, 30, 10}
Now, if we first increase 120 by 10% i.e 120 * 1.1 We get N = 132
again if we increase it by 20% i.e 132 * 1.2 We get N = 158.4
again if we increase it by 30% i.e 158.4 * 1.3 We get N = 205.92
and lastly if we further increase it by 10% i.e 205.92 * 1.1 We get N = 226.51

Now, Percentage change = (226.51 – 120) / 120 = 0.8876
percentage change = 0.8876 * 100 = 88.76 %

这个公式如何运作?
令x为初始值。 A％更改后，x的值变为(x + x * A / 100)
在连续的B％变化之后，x的值变为(x + x * A / 100)+(x + x * A / 100)* B / 100。因此，x的值的增量为x *(A + B + A * B / 100)/ 100。就百分比而言，我们可以说该值增加了(A + B + A * B / 100)％

方法：-

1. 只需在arr [0]和arr [1]之间应用连续更改公式，然后将结果存储在结果变量中
2. 现在，使用上述公式计算result和arr [2]之间的连续变化，并将结果存储在result变量中，依此类推。

   下面是上述方法的实现：

   C++

   // C++ implementation of above approach
   #include 
   using namespace std;
     
   float successiveChange(int arr[], int N)
   {
       float var1, var2, result = 0;
     
       var1 = arr[0];
       var2 = arr[1];
     
       // Calculate successive change of 1st 2 change
       result = var1 + var2 + (float(var1 * var2) / 100);
     
       // Calculate successive change
       // for rest of the value
       for (int i = 2; i < N; i++)
           result = result + arr[i] + (float(result * arr[i]) / 100);
     
       return result;
   }
     
   // Driver code
   int main()
   {
       int arr[] = {10, 20, 30, 10};
       int N = sizeof(arr) / sizeof(arr[0]);
     
       // Calling function
       float result = successiveChange(arr, N);
     
       cout << "Percentage change is = " << result << " %";
       return 0;
   }

   Java

   // Java implementation of above approach
     
   import java.io.*;
     
   class GFG {
     
     
   static float successiveChange(int arr[], int N)
   {
       float var1, var2, result = 0;
     
       var1 = arr[0];
       var2 = arr[1];
     
       // Calculate successive change of 1st 2 change
       result = var1 + var2 + ((var1 * var2) / 100);
     
       // Calculate successive change
       // for rest of the value
       for (int i = 2; i < N; i++)
           result = result + arr[i] + ((result * arr[i]) / 100);
     
       return result;
   }
     
   // Driver code
       public static void main (String[] args) {
           int []arr = {10, 20, 30, 10};
       int N = arr.length;
     
       // Calling function
       float result = successiveChange(arr, N);
     
       System.out.println("Percentage change is = " + result + " %");
       }
   }
   // This code is contributed by shs..

   Python 3

   # Python implementation of above approach
   def successiveChange(arr, N):
     
       result = 0;
     
       var1 = arr[0];
       var2 = arr[1];
     
       # Calculate successive change of 1st 2 change
       result = float(var1 + var2 + 
               (float(var1 * var2) / 100));
         
     
       # Calculate successive change
       # for rest of the value
       for i in range(2, N):
           result = (result + arr[i] +
                    (float(result * arr[i]) / 100));
     
       return result;
     
   # Driver code
   arr = [10, 20, 30, 10];
   N = len(arr) ;
     
   # Calling function
   result = successiveChange(arr, N);
   print ("Percentage change is = %.2f" % 
                          (result), "%");
     
   # This code is contributed 
   # by Shivi_Aggarwal

   C#

   // C# implementation of above approach
   using System;
     
   class GFG {
     
     
   static float successiveChange(int []arr, int N)
   {
       float var1, var2, result = 0;
     
       var1 = arr[0];
       var2 = arr[1];
     
       // Calculate successive change of 1st 2 change
       result = var1 + var2 + ((var1 * var2) / 100);
     
       // Calculate successive change
       // for rest of the value
       for (int i = 2; i < N; i++)
           result = result + arr[i] + ((result * arr[i]) / 100);
     
       return result;
   }
     
   // Driver code
       public static void Main () {
           int []arr = {10, 20, 30, 10};
       int N = arr.Length;
     
       // Calling function
       float result = successiveChange(arr, N);
     
       Console.WriteLine("Percentage change is = " + result + " %");
       }
   }
   // This code is contributed by shs..

   输出：

   Percentage change is = 88.76 %