单利要求本金变成自身Y倍所需的时间

假设在N1年内有一定数量的货币变为自身的T1倍。任务是找到年数，即N2，以使该数量以相同的单利利率成为T2倍。
例子：

Input: T1 = 5, N1 = 7, T2 = 25 
Output: 42
Input: T1 = 3, N1 = 5, T2 = 6 
Output: 12.5 
 

方法：

Let us consider the 1st Example where T1 = 5, N1 = 7, T2 = 25 
Now, Let P principal becomes 5P i.e (T1 * P) then Simple interest received is 4P. 
(As S.I = Amount – P)
Now, in the second case, P has become 25P i.e (T2 * P) then simple interest received is 24P.
Now if we received 4P interest in N1 i.e 7 years then we will get an interest of 24P 
in 7 * 6 years i.e in 42 years. 
 

公式：

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the no. of years
float noOfYears(int t1, int n1, int t2)
{
    float years = ((t2 - 1) * n1 / (float)(t1 - 1));
 
    return years;
}
 
// Driver code
int main()
{
    int T1 = 3, N1 = 5, T2 = 6;
 
    cout << noOfYears(T1, N1, T2);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function to return the no. of years
static float noOfYears(int t1, int n1, int t2)
{
    float years = ((t2 - 1) * n1 / (float)(t1 - 1));
 
    return years;
}
 
// Driver code
public static void main(String[] args)
{
    int T1 = 3, N1 = 5, T2 = 6;
 
    System.out.println(noOfYears(T1, N1, T2));
}
}
 
// This code is contributed by Code_Mech

Python3

# Python3 implementation of the approach
 
# Function to return the no. of years
def noOfYears(t1, n1, t2):
 
    years = (t2 - 1) * n1 / (t1 - 1)
    return years
 
# Driver code
if __name__ == "__main__":
 
    T1, N1, T2 = 3, 5, 6
    print(noOfYears(T1, N1, T2))
     
# This code is contributed
# by Rituraj Jain

C#

// C# implementation for above approach
using System;
 
class GFG
{
     
// Function to return the no. of years
static float noOfYears(int t1, int n1, int t2)
{
    float years = ((t2 - 1) * n1 / (float)(t1 - 1));
 
    return years;
}
 
// Driver code
public static void Main(String[] args)
{
    int T1 = 3, N1 = 5, T2 = 6;
 
    Console.WriteLine(noOfYears(T1, N1, T2));
}
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// Javascript implementaion of the approach
 
// Function to return the no. of years
function noOfYears(t1, n1, t2)
{
    var years = ((t2 - 1) * n1 / (t1 - 1));
 
    return years;
}
 
// Driver code
var T1 = 3, N1 = 5, T2 = 6;
document.write( noOfYears(T1, N1, T2));
 
// This code is contributed by rutvik_56.

输出：

12.5