rgb_scale = 255
cmyk_scale = 100
def rgb_to_cmyk(r,g,b):
if (r == 0) and (g == 0) and (b == 0):
return 0, 0, 0, cmyk_scale
# rgb [0,255] -> cmy [0,1]
c = 1 - r / float(rgb_scale)
m = 1 - g / float(rgb_scale)
y = 1 - b / float(rgb_scale)
min_cmy = min(c, m, y)
c = (c - min_cmy)
m = (m - min_cmy)
y = (y - min_cmy)
k = min_cmy
return c*cmyk_scale, m*cmyk_scale, y*cmyk_scale, k*cmyk_scale
def cmyk_to_rgb(c,m,y,k):
r = rgb_scale*(1.0-(c+k)/float(cmyk_scale))
g = rgb_scale*(1.0-(m+k)/float(cmyk_scale))
b = rgb_scale*(1.0-(y+k)/float(cmyk_scale))
return r,g,b
Answer:
Call tracking system
Explanation:
It helps to measure total amount of visitors per month and based on the volume of visitors, it calculates how many phone numbers should be displayed to show a different number for each clique.
Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
Answer:
l = []
while True:
no = int(input())
if no>0:
l.append(no)
else:
break
print(1)
If this helped consider marking this answer as brainliest. Have a good day.
