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)
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 165 |
2 | maomao90 | 163 |
2 | Um_nik | 163 |
4 | atcoder_official | 161 |
5 | adamant | 160 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | nor | 153 |
9 | Dominater069 | 153 |
How does Counting Inversions work in the following code?
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)
Name |
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