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Slightly more pythonic:
res = sum((i+j)%k == 0 for x, i in enumerate(a) for j in a[x+1:])
Since True == 1
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Divisible Sum Pairs
You are viewing a single comment's thread. Return to all comments →
Slightly more pythonic:
res = sum((i+j)%k == 0 for x, i in enumerate(a) for j in a[x+1:])Since True == 1