Answer: $$98,0,1,0,1.$$ We reason backwards. With a pirate, he keeps the 100 coins: $$100.$$ With two pirates, one vote is enough to reach at least half. The proposer votes for himself, so he can keep everything: $$100,0.$$ With three pirates, at least 2 votes are needed. If the first dies, there would be a case of two pirates: the second would receive 100 and the third 0. The third pirate can be purchased with 1 coin: $$99,0,1.$$ With four pirates, at least 2 votes are needed. If the first one dies, it would be the case of three pirates: $$99,0,1.$$ The third pirate would receive 0 in that scenario, so the proposer offers him 1 coin and keeps the rest: $$99,0,1,0.$$ With five pirates, at least 3 votes are needed. If the first one dies, there would be the case of four pirates: $$99,0,1,0.$$ The third and fifth pirates would receive 0 in that scenario. The proposer already has his own vote, so he buys those two votes with 1 coin each: $$98,0,1,0,1.$$ This distribution gives him enough votes and maximizes what he keeps.