sol1 Algorithm
The sol1 Algorithm, also known as "Squaring the Circle" algorithm, is a mathematical technique that aims to solve the ancient geometrical problem of constructing a square with the same area as a given circle using only compass and straightedge. This algorithm is based on the approximation of the value of Pi (π), which is the ratio of the circumference of a circle to its diameter. The main idea behind the sol1 Algorithm is to find the side length of a square that, when multiplied by itself, gives the same area as that of a circle with a given radius.
The sol1 Algorithm begins by drawing a circle with the desired radius, followed by constructing an inscribed square within the circle. The next step involves dividing the circle's circumference into a number of equal segments, which are then used to create a polygon that approximates the circle. The area of this polygon can be easily calculated using basic trigonometry, and as the number of segments increases, the approximation of the circle's area becomes more accurate. Finally, the side length of the square is determined by finding the square root of the approximated circle's area, and a square with this side length is constructed using a compass and straightedge. Although the sol1 Algorithm provides an approximation to the problem of squaring the circle, it has been proven mathematically impossible to achieve an exact solution using only compass and straightedge due to the transcendental nature of the number π.
"""
Coin sums
Problem 31
In England the currency is made up of pound, £, and pence, p, and there are
eight coins in general circulation:
1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
It is possible to make £2 in the following way:
1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
How many different ways can £2 be made using any number of coins?
"""
def one_pence():
return 1
def two_pence(x):
return 0 if x < 0 else two_pence(x - 2) + one_pence()
def five_pence(x):
return 0 if x < 0 else five_pence(x - 5) + two_pence(x)
def ten_pence(x):
return 0 if x < 0 else ten_pence(x - 10) + five_pence(x)
def twenty_pence(x):
return 0 if x < 0 else twenty_pence(x - 20) + ten_pence(x)
def fifty_pence(x):
return 0 if x < 0 else fifty_pence(x - 50) + twenty_pence(x)
def one_pound(x):
return 0 if x < 0 else one_pound(x - 100) + fifty_pence(x)
def two_pound(x):
return 0 if x < 0 else two_pound(x - 200) + one_pound(x)
def solution(n):
"""Returns the number of different ways can n pence be made using any number of
coins?
>>> solution(500)
6295434
>>> solution(200)
73682
>>> solution(50)
451
>>> solution(10)
11
"""
return two_pound(n)
if __name__ == "__main__":
print(solution(int(str(input()).strip())))
