a = 0
while max(V) > 0:
C = [ 0 ] * max(V)
for x in V:
if x%2==0:
a += C[x//2]
else:
C[x//2] += 1
V = [ x//2 for x in V ]
print(a)
a = 0
while max(V) > 0:
C = [ 0 ] * max(V)
for x in V:
if x%2==0:
a += C[x//2]
else:
C[x//2] += 1
V = [ x//2 for x in V ]
print(a)
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| 1 | Benq | 3792 |
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| Name |
|---|



Consider 2 numbers
and
. Write them down in binary (Assume both of them to be
digit binary number (Incase their length is unequal , pad leading zeroes to the shorter one).
Let
be the minimum index such that
So
is an inversion pair if and only if index of
index of
and
,
and
and
and after that they are both equal so each pair gets counted towards as an inversion exactly once.
So for any