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#!/usr/bin/env lua
--[[
Challenge 120
TASK #2 - Clock Angle
Submitted by: Mohammad S Anwar
You are given time $T in the format hh:mm.
Write a script to find the smaller angle formed by the hands of an analog
clock at a given time.
HINT: A analog clock is divided up into 12 sectors. One sector represents 30
degree (360/12 = 30).
Example
Input: $T = '03:10'
Output: 35 degree
The distance between the 2 and the 3 on the clock is 30 degree.
For the 10 minutes i.e. 1/6 of an hour that have passed.
The hour hand has also moved 1/6 of the distance between the 3 and the 4,
which adds 5 degree (1/6 of 30).
The total measure of the angle is 35 degree.
Input: $T = '04:00'
Output: 120 degree
--]]
function split(inputstr, sep)
if sep == nil then
sep = "%s"
end
local t = {}
for str in string.gmatch(inputstr, "([^"..sep.."]+)") do
table.insert(t, str)
end
return t
end
function parse_time(hhmm)
local a = split(hhmm, ":")
local hh = tonumber(a[1])
local mm = tonumber(a[2])
return hh, mm
end
function clock_angles(hh, mm)
local mm_angle = math.floor(mm*360/60)
local hh_angle = math.floor((hh % 12)*360/12) + math.floor(mm_angle/12)
return hh_angle, mm_angle
end
function clock_angle(hh, mm)
local hh_angle, mm_angle = clock_angles(hh, mm)
local angle = math.abs(hh_angle - mm_angle)
if angle > 180 then angle = 360 - angle end
return angle
end
hh, mm = parse_time(arg[1])
angle = clock_angle(hh, mm)
io.write(angle, "\n")
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