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+17
First, we can split k into power of 2s. Implementing division by 2 is not so hard so I will leave it to you. Then we can notice that its not hard to create a graph that contains exactly 2^k path from 1 to n. For example: Graph with 2 paths.
Graph with 4 paths
Then we can combine these graph together to form k. Graph with 6 paths.
I don't have a exact maximum number of nodes / edges that this requires but it should be well under the limit |
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0
Has the finalists of each track been announced? |
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Can someone help? |
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"change an interval to be equal to some given x" and "calculate the number of positions with a value greater than or equal to y" (sets and fenwick)". Can you briefly explain how to do this? Thanks. |
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Can someone help? |
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Thanks! I got it :)) |
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Auto comment: topic has been updated by Sammmmmmm (previous revision, new revision, compare). |
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+5
No, it's not. Somebody already sent the link. |
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+8
Tysm. That's such a cool solution! |
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Auto comment: topic has been updated by Sammmmmmm (previous revision, new revision, compare). |
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Auto comment: topic has been updated by Sammmmmmm (previous revision, new revision, compare). |
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Auto comment: topic has been updated by Sammmmmmm (previous revision, new revision, compare). |
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Hi, may I ask for the solution for DAY 2 C: Sopsug. Thanks |
