计划从连续年份的复利中找到利率百分比

给定两个整数N1和N2 ，这是连续两年的复利。任务是计算费率百分比。
例子：

Input: N1 = 660, N2 = 720 
Output: 9.09091 %
Input: N1 = 100, N2 = 120 
Output: 20 % 

方法：利率百分比可以使用公式((N2-N1)* 100)/ N1计算，其中N1是某年的复利， N2是下一年的复利。

Let us consider the 1st Example: 
The difference between the Compound interest in the two consecutive years is because of the interest received on the previous year interest. Therefore, 
–> N2 – N1 = N1 * (Rate / 100) 
–> 720 – 660 = 660 * (Rate / 100) 
–> (60 / 660) * 100 = Rate 
–> Rate = (100 / 11) = 9.09% (Approx)  

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the
// required rate percentage
float Rate(int N1, int N2)
{
    float rate = (N2 - N1) * 100 / float(N1);
 
    return rate;
}
 
// Driver code
int main()
{
    int N1 = 100, N2 = 120;
 
    cout << Rate(N1, N2) << " %";
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG
{
 
    // Function to return the
    // required rate percentage
    static int Rate(int N1, int N2)
    {
        float rate = (N2 - N1) * 100 / N1;
 
        return (int)rate;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N1 = 100, N2 = 120;
 
        System.out.println(Rate(N1, N2) + " %");
    }
}
 
// This code has been contributed by 29AjayKumar

Python 3

# Python 3 implementation of the approach
 
# Function to return the
# required rate percentage
def Rate( N1, N2):
    rate = (N2 - N1) * 100 // (N1);
 
    return rate
 
# Driver code
if __name__ == "__main__":
    N1 = 100
    N2 = 120
 
    print(Rate(N1, N2) ," %")
 
# This code is contributed by ChitraNayal

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to return the
    // required rate percentage
    static int Rate(int N1, int N2)
    {
        float rate = (N2 - N1) * 100 / N1;
 
        return (int)rate;
    }
 
    // Driver code
    static public void Main ()
    {
        int N1 = 100, N2 = 120;
 
        Console.WriteLine(Rate(N1, N2) + " %");
    }
}
 
// This code has been contributed by ajit.

Javascript

 
// javascript implementation of the approach
 
    // Function to return the
    // required rate percentage
    function Rate(N1 , N2) {
        var rate = (N2 - N1) * 100 / N1;
 
        return parseInt( rate);
    }
 
    // Driver code
     
        var N1 = 100, N2 = 120;
 
        document.write(Rate(N1, N2) + " %");
 
// This code contributed by Rajput-Ji
 

输出：

20 %

时间复杂度： O(1)

辅助空间： O(1)