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.
"""
Number letter counts
Problem 17
If the numbers 1 to 5 are written out in words: one, two, three, four, five,
then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in
words, how many letters would be used?
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20
letters. The use of "and" when writing out numbers is in compliance withBritish
usage.
"""
def solution(n):
"""Returns the number of letters used to write all numbers from 1 to n.
where n is lower or equals to 1000.
>>> solution(1000)
21124
>>> solution(5)
19
"""
# number of letters in zero, one, two, ..., nineteen (0 for zero since it's
# never said aloud)
ones_counts = [0, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8]
# number of letters in twenty, thirty, ..., ninety (0 for numbers less than
# 20 due to inconsistency in teens)
tens_counts = [0, 0, 6, 6, 5, 5, 5, 7, 6, 6]
count = 0
for i in range(1, n + 1):
if i < 1000:
if i >= 100:
# add number of letters for "n hundred"
count += ones_counts[i // 100] + 7
if i % 100 != 0:
# add number of letters for "and" if number is not multiple
# of 100
count += 3
if 0 < i % 100 < 20:
# add number of letters for one, two, three, ..., nineteen
# (could be combined with below if not for inconsistency in
# teens)
count += ones_counts[i % 100]
else:
# add number of letters for twenty, twenty one, ..., ninety
# nine
count += ones_counts[i % 10]
count += tens_counts[(i % 100 - i % 10) // 10]
else:
count += ones_counts[i // 1000] + 8
return count
if __name__ == "__main__":
print(solution(int(input().strip())))
