John has some amount of money of which he wants to deposit a part `f0` to the bank at the beginning
of year `1`. He wants to withdraw each year for his living an amount `c0`.

Here is his banker plan:

- deposit `f0` at beginning of year 1
- his bank account has an interest rate of `p` percent per year, constant over the years
- John can withdraw each year `c0`, taking it whenever he wants in the year; he must take account of an inflation of `i` percent per year in order to keep his quality of living. `i` is supposed to stay constant over the years.
- all amounts f0..fn-1, c0..cn-1 are truncated by the bank to their integral part
- Given f0, `p`, c0, `i` 
the banker guarantees that John will be able to go on that way until the `nth` year.

# Example:

```
f0 = 100000, p = 1 percent, c0 = 2000, n = 15, i = 1 percent
```

```
beginning of year 2 -> f1 = 100000 + 0.01*100000 - 2000 = 99000;  c1 = c0 + c0*0.01 = 2020 (with inflation of previous year)
```

```
beginning of year 3 -> f2 =  99000 + 0.01*99000 - 2020  = 97970;  c2 = c1 + c1*0.01 = 2040.20 
(with inflation of previous year, truncated to 2040)
```

```
beginning of year 4 -> f3 =  97970 + 0.01*97970 - 2040  = 96909.7 (truncated to 96909); 
c3 = c2 + c2*0.01 = 2060.4 (with inflation of previous year, truncated to 2060)
```
and so on...

John wants to know if the banker's plan is right or wrong.
Given parameters `f0, p, c0, n, i` build a function `fortune` which returns `true` if John can make a living until the `nth` year
and `false` if it is not possible.

# Some cases:
```
fortune(100000, 1, 2000, 15, 1) -> True
fortune(100000, 1, 10000, 10, 1) -> True
fortune(100000, 1, 9185, 12, 1) -> False

For the last case you can find below the amounts of his account at the beginning of each year:
100000, 91815, 83457, 74923, 66211, 57318, 48241, 38977, 29523, 19877, 10035, -5
```
f11 = -5 so he has no way to withdraw something for his living in year 12.

> **Note:** Don't forget to convert the percent parameters as percentages in the body of your function: if a parameter percent is 2 you have to convert it to 0.02.

Difficulty: EASY
Tags: Mathematics

Starter code:
def fortune(f0, p, c0, n, i):
	

Solutions:
def fortune(f, p, c, n, i):
	for _ in range(n - 1):
		f = int(f * (100 + p) / 100 - c)
		c = int(c * (100 + i) / 100)
		if f < 0:
			return False
	return True


def fortune(money, interest, withdraw, years, inflation):
	interest = 1 + interest / 100
	inflation = 1 + inflation / 100
	for _ in range(years - 1):
		money = int(money * interest - withdraw)
		if money < 0:
			return False
		withdraw *= inflation
	return True


import math

def fortune(f0, p, c0, n, i):
	f0 = math.trunc(f0 * (1 + p / 100) - c0)
	c0 = c0 * (1 + i / 100)
	if n == 2:
		return True if f0 >= 0 else False
	else:
		return fortune(f0, p, c0, n - 1, i)


def fortune(f0, p, c0, n, i):
	f = f0
	c = c0
	for year in range(n - 1):
		f = int(f + f * (p / 100)) - int(c)
		c = int(c + c * (i / 100))
	return f >= 0


def fortune(f0, p, c0, n, i):
	for t in range(n - 1):
		f0 = int(f0 + f0 * p / 100 - c0)
		c0 = int(c0 + c0 * i / 100)
	if f0 >= 0:
		return True
	return False


def fortune(f0, p, c0, n, i):
	year_checker = 0
	fn = f0
	cn = c0
	while year_checker < n - 1:
		fn = int(fn + fn * (0.01 * p) - cn)
		cn = int(cn + cn * (0.01 * i))
		year_checker = year_checker + 1
	if fn < 0:
		return False
	return True


def fortune(f0, p, c0, n, i):
	year = 1
	while f0 > 0 and year < n:
		f0 = f0 * (1 + p / 100) - c0
		f0 = int(f0)
		c0 = c0 * (1 + i / 100)
		c0 = int(c0)
		year += 1
	return True if f0 >= 0 else False


def fortune(f0, p, c0, n, i):
	print(f0, p, c0, n, i)
	for _ in range(n - 1):
		f0 = f0 * (100 + p) // 100 - c0
		c0 = c0 * (100 + i) // 100
	return f0 >= 0


def fortune(f0, p, c0, n, i):
	inf = c0
	money = f0
	for j in range(n - 1):
		money = int(money + money * (p / 100) - inf)
		inf = int(inf + inf * (i / 100))
	return True if money >= 0 else False


fortune = F = lambda f, p, c, n, i: F(f * (100 + p) // 100 - c, p, c * (100 + i) // 100, n - 1, i) if n > 1 else f >= 0