In this Leetcode Roy and LEDs problem solution Its Diwali time and there are LED series lights everywhere. Little Roy got curious about how LED lights work. He noticed that in one single LED Bulb there are 3 LED lights, namely Red, Green, and Blue. The state of the bulb at any moment is the sum of Red, Green, and Blue LED lights.

The bulb works as follows:

Roy took out all the LEDs and found that Red LED stays ON for R seconds, Green LED stays ON for G seconds and Blue LED stays ON for B seconds. Similarly, they stay OFF for the same respective R, G, B number of seconds. (Initially, all the LEDs are OFF. See Sample Test Case Explanation for better understanding)

Roy has one query for you, given the total number of seconds T, find the number of seconds Red, Green, Blue, Yellow, Cyan, Magenta, White, Black(no light) lights are visible.

## HackerEarth Roy and LEDs problem solution.

```def led():
T,R,G,B = map(int,raw_input().split())
red = *1000001
green = *1000001
blue = *1000001
colors = *8
r = 0
rp = 0
g = 0
gp = 0
b = 0
bp = 0
for i in xrange(T):
if i >= R + 2*R*r and i < 2*R + 2*R*r:
red[i] = 1
rp += 1
if rp%R == 0:
r += 1
if i >= G + 2*G*g and i < 2*G + 2*G*g:
green[i] = 1
gp += 1
if gp%G == 0:
g += 1
if i >= B + 2*B*b and i < 2*B + 2*B*b:
blue[i] = 1
bp += 1
if bp%B == 0:
b += 1
for i in xrange(T):
if red[i] == 1 and green[i] == 0 and blue[i] == 0:   #red
colors += 1
elif red[i] == 0 and green[i] == 1 and blue[i] == 0: #green
colors += 1
elif red[i] == 0 and green[i] == 0 and blue[i] == 1: #blue
colors += 1
elif red[i] == 1 and green[i] == 1 and blue[i] == 0: #yellow
colors += 1
elif red[i] == 0 and green[i] == 1 and blue[i] == 1: #cyan
colors += 1
elif red[i] == 1 and green[i] == 0 and blue[i] == 1: #magenta
colors += 1
elif red[i] == 1 and green[i] == 1 and blue[i] == 1: #white
colors += 1
else:                                                #black
colors += 1
print ' '.join(str(x) for x in colors)
led()```