First, the user is asking about the pirate game, a classic game theory problem. It involves five pirates who must distribute 100 gold coins, with each pirate proposing a distribution in order (from 1 to 5), and others voting. The proposal needs more than half the votes to pass; otherwise, the proposer is thrown overboard.

The key points:

• Pirates are labeled 1 to 5, with pirate 1 proposing first.
• Voting: After a proposal, all pirates vote (including the proposer? I think yes, based on standard versions).
• Majority needed: More than half agree. Since there are 5 pirates initially, majority means at least 3 votes (since 3 > 2.5, and it must be more than half).
• If rejected, the proposer is killed, and the next pirate proposes with fewer pirates.
• Pirates are rational and prioritize: 1. Survival, 2. Maximizing gold, 3. Throwing others overboard if indifferent.

I need to solve this from a multi-step game perspective, meaning I should work backwards from the end because it's a sequential game with perfect information.

Start from the smallest number of pirates and build up.

Case with 1 pirate:

• If only one pirate, say pirate 5, he takes all 100 coins. No vote needed.

Case with 2 pirates:

• Pirates: 4 and 5. Pirate 4 proposes.
• Votes: Pirate 4 and pirate 5.
• Majority needed: More than half of 2 is 1, but since it's "more than half", for 2 pirates, half is 1, so more than half means 2 votes? I need to clarify.

Standard interpretation: For n pirates, majority is strictly more than n/2 votes.

• For 2 pirates: n=2, n/2=1, so more than 1 means at least 2 votes. So, pirate 4 needs both votes to pass? But if he proposes, he votes for himself, so he needs one more vote.

Typically, in this game, the proposer votes for their own proposal, so:

• With 2 pirates: Pirate 4 proposes. He votes yes. Pirate 5 votes. If pirate 5 votes no, then only one yes vote, which is not more than half (since half of 2 is 1, and more than half requires >1, so 2 votes needed). So, it requires both to agree? That might not be right.

I think I recall that for even numbers, it's tricky. Let me confirm the standard setup.

Upon second thought, in most versions, "more than half" means at least ceil(n/2) + 0.5, but practically, for n pirates, it requires at least floor(n/2) + 1 votes or something. Let's think.

For n=5: half is 2.5, more than half is at least 3 votes.

For n=4: half is 2, more than half is at least 3 votes? No, more than 2 means 3 or more, but for 4 pirates, 3 votes are needed since 2 is exactly half, not more.

Standard rule: the proposal passes if the number of yes votes is strictly greater than no votes. Since there might be abstentions, but in this game, pirates vote yes or no, so it's based on votes cast.

In this problem, it says "投票要超过半数同意", which means "votes must exceed half agreed". So, for n pirates, it needs votes > n/2.

• For n=5, needs >2.5, so at least 3.
• For n=4, needs >2, so at least 3 votes.
• For n=3, needs >1.5, so at least 2 votes.
• For n=2, needs >1, so at least 2 votes? >1 means 2 or more, but since there are only 2 pirates, it must be both agree.
• For n=1, no vote.

I think for n=2, it requires both to agree, which makes sense because if only one agrees, it's not more than half.

But let's proceed with standard game theory approach.

Work backwards:

Start with 2 pirates: Pirate 4 and Pirate 5.

• Pirate 4 proposes a distribution.
• He needs votes >1 (since n=2, n/2=1, >1 means 2 votes).
• So, he must get both votes to pass.
• Pirate 4 will vote yes for any proposal since he wants to survive.
• Pirate 5 will vote yes only if he gets at least 1 coin; otherwise, he might vote no to kill pirate 4 and be alone, but since he gets nothing if pirate 4 is killed? If proposal is rejected, pirate 4 is killed, and pirate 5 is alone, so pirate 5 gets all 100.
• Therefore, pirate 5 would prefer to vote no if he gets less than 100, because if proposal rejected, he gets 100 alone.
• But pirate 4 needs pirate 5 to vote yes, so he must offer pirate 5 something to make him vote yes.
• If pirate 4 offers (0,100) to (4,5), pirate 5 gets 100, so he votes yes.
• Pirate 4 gets 0, but he survives.
• Pirate 4 could offer (99,1) or something, but since pirate 5 gets 100 if he votes no, he will only vote yes if he gets at least 100? That can't be right.

Mistake: If pirate 4 proposes, and it's rejected, pirate 4 is killed, and then only pirate 5 remains, so pirate 5 gets 100.

For pirate 5:

• If he votes yes, he gets whatever is offered in the proposal.
• If he votes no, and the proposal is rejected (which requires both votes? No.

With 2 pirates:

• Pirate 4 proposes. Let’s say he offers X to himself and Y to pirate 5, X+Y=100.
• Pirate 4 votes yes (since he wants it to pass to survive).
• Pirate 5 votes. If pirate 5 votes yes, proposal passes, he gets Y.
• If pirate 5 votes no, then only one yes vote (pirate 4), which is not >1 (since 1 <=1), so it fails, pirate 4 is killed, pirate 5 alone gets 100.
• So, pirate 5 will vote yes only if Y >=100? But Y <=100, so only if Y=100, he gets 100, same as if he votes no. But pirates are rational, and they prefer not to throw others overboard if indifferent, but in terms of gold, if Y=100, he gets 100 either way, so he might vote yes to save pirate 4 or no to kill him.

The priority: survival first, then maximize gold, then minimize others' gold or throw overboard.

For pirate 5:

• If he votes no, and it fails, he gets 100 gold alone.
• If he votes yes, he gets Y gold.
• So, he will vote yes only if Y > 100? Impossible, or if Y >=100, but since X+Y=100, Y=100 only if X=0.

But if X=0, Y=100, pirate 5 gets 100 if he votes yes, or 100 if he votes no. Since he is indifferent in gold, but he might prefer to vote no to kill pirate 4, as per the third priority.

In standard game theory for this problem, when indifferent, pirates will vote to save the proposer to avoid risk or something, but I think the common assumption is that if indifferent, they vote yes to pass the proposal.

Let's recall the standard solution.

Typically, for 2 pirates:

• Pirate 4 can propose to give 0 to himself and 100 to pirate 5.
• Pirate 4 votes yes.
• Pirate 5: if he votes yes, he gets 100, and proposal passes.
• If he votes no, proposal fails (since only one yes vote), pirate 4 killed, pirate 5 gets 100 alone.
• So he is indifferent. But to maximize gold, he gets 100 either way. The third priority is to throw others overboard, so he might vote no to kill pirate 4.
• However, in most versions, it's assumed that if indifferent, they vote in a way that passes the proposal to avoid unnecessary death, or for simplicity, they vote yes.

But to be precise, let's look for a reliable source in my knowledge.

I remember that in the standard pirate game, for 2 pirates, pirate 4 gives 100 to pirate 5 and 0 to himself, and pirate 5 votes yes because if he votes no, he gets nothing extra, but since he gets 100 anyway, and voting no might not change his gold, but the proposal failing means he is alone, so he is fine, but he has no incentive to vote no.

The key is that pirates prefer more gold, and if gold is same, they prefer to save others or kill, but the third priority is to throw overboard if possible.

From the problem: "海盗是理性的，优先考虑自身生存，然后最大化金币，最后如果无差别，就希望把别人扔下海。"

So, priorities:

1. Survival
2. Maximize gold
3. If indifferent, throw others overboard (i.e., vote to kill if possible).

For 2 pirates:

• Pirate 4 proposes. He must get both votes for it to pass, since >1 vote needed.
• Pirate 4 votes yes to survive.
• Pirate 5: if offered Y, he votes yes if Y > what he gets if rejected, or if equal, he might vote no to kill.
• If rejected, pirate 5 gets 100 alone.
• So, if Y < 100, pirate 5 prefers to vote no, get 100, and kill pirate 4.
• If Y = 100, he gets 100 either way, but since indifferent, he would vote no to throw pirate 4 overboard.
• Therefore, pirate 5 will only vote yes if Y > 100, impossible, so he always votes no.
• Then, only one yes vote, not >1, so proposal fails, pirate 4 killed, pirate 5 gets 100.

But that means with 2 pirates, no proposal can pass? That can't be right.

For n=2, if pirate 4 proposes anything, pirate 5 votes no to kill him and get 100, so it always fails. Then pirate 4 is killed, and pirate 5 gets 100 alone.

But that means pirate 4 cannot propose anything to save himself, so he dies, and pirate 5 benefits.

In that case, with 2 pirates, pirate 4 is always killed if he proposes, so it's better for him not to be in that position, but in the sequence, he has to propose.

For n=2, the proposal will always fail because pirate 5 can vote no to kill pirate 4 and get all.

But then, when we have more pirates, we need to consider that.

I think I have a mistake. Let's check online or standard knowledge.

Upon recollection, in the standard 5 pirate problem, for 2 pirates, pirate 4 can propose to give 0 to himself and 100 to pirate 5, and pirate 5 votes yes because if he votes no, he still gets 100, but by voting yes, he gets 100 and saves pirate 4, which might be preferred since throwing overboard is only if indifferent and no cost, but here voting yes saves pirate 4 at no cost, so he might vote yes.

But the priority is: if indifferent in gold, they want to throw others overboard. So if gold is same, they prefer to vote no to kill.

Let's assume that.

For 2 pirates:

• If pirate 4 offers (0,100), pirate 5 gets 100 if yes, or 100 if no. Since gold same, he votes no to kill pirate 4.
• Then only one yes vote, not >1, so fails, pirate 4 killed, pirate 5 gets 100.
• If pirate 4 offers (1,99), pirate 5 gets 99 if yes, or 100 if no, so he prefers to vote no to get 100.
• Similarly for any offer less than 100 to pirate 5, he votes no.
• Only if he offers 100 to pirate 5, but pirate 5 still votes no to kill.
• So no proposal can pass, pirate 4 always dies.

But that can't be, because in the 5 pirate problem, it works with backward induction.

Perhaps for n=2, "more than half" with 2 pirates: half is 1, more than half is 2 votes, so both must agree.

But since pirate 5 can force a no, it fails.

However, in many sources, it's assumed that for n=2, pirate 4 gives 100 to pirate 5, and pirate 5 votes yes because he is rational and gets the gold without killing, but according to the priorities, if he is indifferent, he wants to kill.

To resolve this, I think for the sake of the game, we assume that when indifferent in gold, they vote in a way that is neutral, but the standard solution has them voting to pass.

Let's look for the correct approach.

I recall that in the pirate game, pirates will accept an offer if they get at least as much as they would in the next round, considering their payoff if the proposer is killed.

For 2 pirates:

• If proposal rejected, pirate 4 killed, pirate 5 gets 100.
• So for pirate 5, his alternative is 100.
• He will vote yes only if he gets at least 100 in the proposal.
• Since total is 100, he must get 100, so pirate 4 offers (0,100).
• Pirate 5 gets 100, which is equal to his alternative, so he is indifferent.
• Since he is indifferent, and the third priority is to throw others overboard if possible, he will vote no to kill pirate 4.
• Therefore, proposal fails, pirate 4 killed, pirate 5 gets 100.

But then with 2 pirates, the proposal always fails, and pirate 4 dies.

Now for 3 pirates: pirates 3,4,5.

Pirate 3 proposes.

He needs votes >1.5, so at least 2 votes (since 2>1.5).

He will vote for himself, so he needs one more vote.

If proposal rejected, pirate 3 killed, then only pirates 4 and 5 left, and as above, with 2 pirates, proposal fails, pirate 4 killed, pirate 5 gets 100.

So in the subgame with 2 pirates, pirate 4 gets 0, pirate 5 gets 100.

Back to 3 pirates:

• Pirate 3 proposes.
• He needs one more vote.
• Pirate 4: if proposal rejected, he gets 0 (since he dies in the next round).
• Pirate 5: if proposal rejected, he gets 100.

So pirate 4 will accept any offer with at least 1 gold, because if rejected, he gets 0.

Pirate 5 will accept only if offered at least 100, but total is 100, so not possible.

Pirate 3 can bribe pirate 4 with 1 gold, and give 99 to himself, and 0 to pirate 5.

Then:

• Pirate 3 votes yes.
• Pirate 4: offered 1, which is better than 0 if rejected, so votes yes.
• Pirate 5: offered 0, but if rejected, he gets 100, so he votes no.
• Votes: two yes (3 and 4), one no (5), so 2 > 1.5, passes.

So proposal: (99,1,0) for (3,4,5).

Pirate 5 gets 0, but he wanted to vote no to get 100, but since the proposal passes, he gets 0.

Now for 4 pirates: pirates 2,3,4,5.

Pirate 2 proposes.

Needs votes >2, so at least 3 votes (since for n=4, >2 means at least 3).

He votes for himself, so needs two more votes.

If proposal rejected, pirate 2 killed, then pirates 3,4,5 left, and from above, with 3 pirates, pirate 3 gets 99, pirate 4 gets 1, pirate 5 gets 0.

So payoffs if rejected: pirate 3:99, pirate 4:1, pirate 5:0.

Pirate 2 needs to get two of them to vote yes.

He can target those who get less in the next round.

Pirate 5 gets 0 if rejected, so he will accept any positive offer.

Pirate 4 gets 1, so he needs at least 1 to be indifferent or better.

Pirate 3 gets 99, so he needs at least 99 to accept.

Pirate 2 can offer 1 to pirate 5, who will accept since 1 > 0.

He can offer 1 to pirate 4, but pirate 4 gets 1 already, so if offered 1, he is indifferent; since indifferent, he might vote no to throw overboard.

Similarly, pirate 3 would need 99 to be indifferent, but pirate 2 can offer less.

To save cost, pirate 2 should offer to those who are cheaper to bribe.

Pirate 5: will accept any offer >0, since he gets 0 otherwise.

Pirate 4: gets 1 if rejected, so he will accept only if offered >1, or if =1, he is indifferent and might vote no.

Pirate 3: needs >99 to accept, but not possible or expensive.

So pirate 2 can offer to pirate 5 with 1 gold, and to someone else.

He needs two votes besides himself.

He can offer 0 to pirate 3, 0 to pirate 4, and 1 to pirate 5, and 99 to himself.

Then:

• Pirate 2 votes yes.
• Pirate 3: offered 0, but if rejected, he gets 99, so he prefers to vote no.
• Pirate 4: offered 0, if rejected he gets 1, so he prefers to vote no (since 0<1).
• Pirate 5: offered 1, if rejected he gets 0, so he votes yes (1>0).
• Votes: pirate 2 yes, pirate 5 yes, pirate 3 no, pirate 4 no. So two yes, two no. For n=4, >2 votes needed, 2 is not greater than 2, so it fails? 2 > 2? No, 2 is not greater than 2, so it fails if tied.

"超过半数" means strictly more than half.

For n=4, half is 2, so >2 means at least 3 votes.

With two yes, it's not enough, so fails.

Pirate 2 needs three yes votes.

He votes yes, so he needs two more.

He must get two others to vote yes.

Pirate 5 will vote yes if offered at least 1 (since he gets 0 otherwise).

Pirate 4 gets 1 if rejected, so to get his vote, need to offer more than 1, say 2.

Pirate 3 gets 99, so expensive.

So pirate 2 can offer 2 to pirate 4 and 1 to pirate 5, and 97 to himself, 0 to pirate 3.

Then:

• Pirate 2 yes
• Pirate 4: offered 2 >1, so votes yes
• Pirate 5: offered 1 >0, so votes yes
• Pirate 3: offered 0 <99, votes no
• Votes: three yes (2,4,5), one no (3), so 3>2, passes.

Pirate 4 was offered 2, which is better than 1, so he votes yes.

If pirate 2 offers only 1 to pirate 4, pirate 4 gets 1, same as if rejected, so indifferent. Since indifferent, he might vote no to throw overboard, as per priority.

To ensure his vote, offer more than what he gets.

Similarly, pirate 5 gets 0, so offer 1 is sufficient.

So proposal: pirate 2:97, pirate 3:0, pirate 4:2, pirate 5:1? Total 97+0+2+1=100.

But pirate 4 gets 2, which is more than 1.

Yes.

Pirate 2 could offer pirate 5 with 1, and pirate 3 with something, but pirate 3 needs 99 to be indifferent, which is costly, so better to bribe cheaper pirates.

Now for 5 pirates: pirates 1,2,3,4,5.

Pirate 1 proposes.

Needs votes >2.5, so at least 3 votes.

He votes for himself, so needs two more votes.

If proposal rejected, pirate 1 killed, then pirates 2,3,4,5 with pirate 2 proposing, and from above, with 4 pirates, pirate 2 gets 97, pirate 3 gets 0, pirate 4 gets 2, pirate 5 gets 1.

So payoffs if rejected: pirate 2:97, pirate 3:0, pirate 4:2, pirate 5:1.

Pirate 1 needs to get two of them to vote yes.

He can target those who get less.

Pirate 3: gets 0 if rejected, so he will accept any positive offer, say 1 gold.

Pirate 4: gets 2, so he needs more than 2 to accept.

Pirate 5: gets 1, so needs more than 1.

Pirate 2: gets 97, so needs at least 97 to accept, expensive.

Pirate 1 can bribe pirate 3 and pirate 5.

Pirate 3: offer 1, since he gets 0, so he votes yes for 1.

Pirate 5: gets 1 if rejected, so to get his vote, need to offer more than 1, say 2.

Then pirate 1 can take 97 for himself, offer 1 to pirate 3, 2 to pirate 5, and 0 to others.

Total: 97 + 0 + 1 + 0 + 2? Pirates: 1,2,3,4,5.

Distribution: pirate 1:97, pirate 2:0, pirate 3:1, pirate 4:0, pirate 5:2.

But pirate 4 is offered 0, but if rejected, he gets 2, so he prefers to vote no since 0<2.

Similarly, pirate 2 offered 0 <97, votes no.

Pirate 3 offered 1 >0, votes yes.

Pirate 5 offered 2 >1? 2>1, so he should vote yes.

Pirate 1 votes yes.

Votes: pirate 1 yes, pirate 2 no, pirate 3 yes, pirate 4 no, pirate 5 yes.

So three yes (1,3,5), two no (2,4), so 3>2.5, passes. More than half of 5 is 2.5, 3>2.5, yes.

Pirate 5 was offered 2, which is greater than 1 he gets if rejected, so he strictly prefers yes.

Pirate 3 offered 1 >0, votes yes.

Pirate 4 offered 0 <2, so votes no.

But pirate 1 has the votes: himself, pirate 3, and pirate 5.

He could offer pirate 4 instead, but pirate 4 needs more than 2, say 3, which is more expensive than offering 2 to pirate 5.

Offering 1 to pirate 3 and 2 to pirate 5 costs 3, same as offering 3 to pirate 4, but then pirate 5 might not vote, but with pirate 3 and pirate 5, he has the votes.

In this case, it works.

He could offer to pirate 4 with 3, and pirate 3 with 1, but then pirate 5 gets 1 if rejected, and offered less, might vote no, but he doesn't need pirate 5 if he has pirate 4 and pirate 3.

But offering 3 to pirate 4 and 1 to pirate 3 costs 4, whereas previous costs 3 (1+2), so better to offer less.

97 to himself, 1 to pirate 3, 2 to pirate 5, 0 to pirate 2 and pirate 4.

Total: 97+0+1+0+2=100.

Yes.

Pirate 4 is offered 0, gets 2 if rejected, so votes no.

But since pirate 1 has three votes, it passes.

Now, is there a way to save gold? For example, if he offers pirate 5 with 1, but pirate 5 gets 1 if rejected, so if offered 1, he is indifferent; since indifferent, he might vote no to throw overboard.

Similarly for pirate 3.

In this case, for pirate 5: if offered exactly 1, he gets 1 if yes, or 1 if rejected? If rejected, he gets 1 in the next round? With 4 pirates, he gets 1, as above.

In the 4-pirate case, pirate 5 gets 1.

So if pirate 1 offers 1 to pirate 5, pirate 5 gets 1 if he votes yes, or if he votes no and proposal rejected, he still gets 1 in the next round.

So he is indifferent in gold.

Since indifferent, he will vote no to throw pirate 1 overboard, as per the third priority.

Similarly, pirate 3: if offered 1, he gets 1 if yes, or 0 if rejected? If rejected, with 4 pirates, pirate 3 gets 0, as above.

In 4-pirate case, pirate 3 gets 0.

So if offered 1, he gets 1 >0, so he strictly prefers yes.

For pirate 5: if offered 1, he gets 1 if yes, or 1 if rejected (since in next round he gets 1), so gold same, indifferent, so he votes no to kill.

Therefore, if pirate 1 offers 1 to pirate 5, pirate 5 votes no because indifferent and wants to throw overboard.

To get pirate 5's vote, he must offer more than what he gets, so more than 1, say 2.

Similarly, for pirate 3, offering 1 is sufficient since 1>0.

Pirate 4 gets 2, so if offered less, he votes no, but not needed.

So optimal: pirate 1 offers himself 98? Let's calculate.

He needs to offer to pirate 3: at least 1, but since 1>0, he accepts.

To pirate 5: must offer more than 1, so at least 2? Since if offered 2, 2>1, strictly better, so votes yes.

If offered exactly 1, indifferent, votes no.

So offer 2 to pirate 5.

Then to pirate 3: offer 1.

Pirate 2 and 4: offer 0.

Himself: 100 -1 -2 =97? 97 to himself.

97+1+2=100, yes.

As before.

He could offer pirate 4 instead.

Pirate 4 gets 2 if rejected, so offer more than 2, say 3.

Then pirate 3 offer 1.

Total cost 4, himself 96, worse.

Or offer pirate 2, but he gets 97, expensive.

So best is to offer pirate 3 with 1 and pirate 5 with 2.

Thus, proposal: pirate 1:97, pirate 2:0, pirate 3:1, pirate 4:0, pirate 5:2.

Now, to confirm the votes:

• Pirate 1: votes yes

• Pirate 2: offered 0 <97, so votes no

• Pirate 3: offered 1 >0, so votes yes

• Pirate 4: offered 0 <2, so votes no

• Pirate 5: offered 2 >1, so votes yes

Yes votes: 1,3,5 → three yes, which is more than half (2.5), so passes.

Pirate 1 maximizes his gain at 97.

But the user asked for the first pirate, which is pirate 1.

In the sequence, pirate 1 proposes first.

So allocation: pirate 1 gets 97, and bribes pirate 3 and pirate 5.

Now, to write the response.

The user message is in Chinese, so final response should be in Chinese.

User said: "5个海盗抢得100枚金币：..." and the question in Chinese.

So reply in Chinese.

But first, in the reasoning, I used English.

Now, for the final answer.

Pirate 1 should propose: himself 97, pirate 2: 0, pirate 3: 1, pirate 4: 0, pirate 5: 2.

To state it clearly.