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# approach for BEERUS

 1 Can anyone discuss their approach for this problem from this contest This question is marked "community wiki". asked 13 Jun '18, 11:15 4★sonu_628 420●8 accept rate: 10%

 3 The only catch in the problem was that (node(i) OR node(j)) + (node(i) AND node(j)) = node(i) + node(j).The rest is easy to figure out. answered 13 Jun '18, 12:13 6★smartnj 57●3 accept rate: 50% Can you tell me where did you learn this property that A|B + A&B = A + B (14 Jun '18, 12:48) ay23064★
 1 From this I see that MST is a star with center in lightweight node. And weight MST = Sum{node_i} + (n-2)*node_0, node_0 - is lightweight node. Sum {node_i} - easy to get. But how to get weight of lightweight node? answered 13 Jun '18, 19:51 126●4 accept rate: 5% 2 If you have $\displaystyle\sum_{i}^{} node_i$ it should be straightforward to obtain the weights for each node. $node_i = \frac{1}{n} * \left( gohan[i] + trunks[i] - \displaystyle\sum_{i}^{} node_i\right)$. Now just loop over all nodes and obtain the maximum. (13 Jun '18, 21:58)
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question asked: 13 Jun '18, 11:15

question was seen: 281 times

last updated: 14 Jun '18, 12:48