There are 5 robbers who divide the spoils, assuming a total of 100 dollars 10

Thinking about it backwards, if the first three people are dead, then the fourth person will die no matter what plan he uses, so if he wants to save his life, he must either take 0 or save the third person.

Looking at the third person, if the first two are over, as long as he proposes a plan, No. 4 will definitely support him, and as long as No. 5 is against it, he will be over, so either take 0 or support No. 2's plan.

Looking at No. 1 again, No. 2 will definitely oppose him, because as long as No. 1 dies, No. 2's plan will definitely be supported by No. 3 and No. 4.

Looking at No. 5 again, he obviously doesn't want No. 2 to be distributed, because even if No. 2 is 100,0,0,0 so that he can pass, he won't be able to get any of them.

So for No. 1, he only needs to give No. 3, No. 4, and No. 5 1 each, then their situation is better than the situation later, then they will support him and make the plan pass.

Backwards method. If the first three people are dead, then No. 4 will die no matter what plan he uses, and he wants to save his life, either proposing a plan or agreeing to No. 3's plan.

No. 3, if the first two are over, no matter what plan he proposes, No. 4 will definitely support him, so he can mention: No., to get more than half of the support, you must give No. 4 or No. 5 more money than No. 3, No. 4 can give him 0; If No. 5 gets more than the No. 3 plan, he will agree to No. 2, so No. 2 will raise the number, and if you want to get more than half, you must let someone in No. 3, No. 4, and No. 5 get more money than No. 2's plan, because No. 4 considers saving his life and gives No. 4 0, No. 4 will agree, so as long as you consider letting No. 3 and No. 5 get more money than No. 2's plan. Therefore, the number 1 will mention:

No one gets a penny.

The distribution plan is that 3 of the 5 thieves died, and the remaining 2 were divided equally.

I don't know how to tell me about you.

Pushing back forward, if robbers 1 and 3 feed sharks, and only 4 and 5 are left, 5 must vote against letting 4 feed sharks in order to swallow all the gold coins. Therefore, No. 4 can only save his life if he supports No. 3. Knowing this, No. 3 will propose a (100,0,0) distribution plan, and will not pull out a dime for No. 4 and No. 5 and classify all the gold coins as own, because he knows that No. 4 will get nothing but will still vote in favor, plus his own vote, his plan can be passed.

However, when 2 learns about 3's plan, he will propose a plan (98,0,1,1), that is, give up 3 and give 4 and 5 a gold coin each. Since the scheme is more advantageous for No. 4 and No. 5 than when they are assigned to No. 3, they will support him and do not want him to be out and assigned by No. 3. In this way, the number 2 will take 98 gold coins.

However, the plan of the 2nd will be understood by the 1st, and the 1st will propose (97,0,1,2,0) or (97,0,1,0,2) that is, give the 2nd and give the 3rd brother a gold coin, and the 4th (or 5th) 2 gold coins. Since the 1st plan is better for the 3rd and 4th (or 5th) than the 2nd allocation, they will vote for the 1st, and with the 1st's own vote, the 1st's plan will be approved, and the 97 gold coins will easily fall into the pocket. This is undoubtedly the plan that No. 1 can get the most bang for its buck.

Is it the correct proposal: "No. 1 takes 98 by himself, No. 2 does not give it, No. 3 gives 1 seed, No. 4 1 pill, No. 5 does not give it!" ”

Why? This is a typical battle problem - how to follow the rules and work backwards from the final goal to the most advantageous first move.

Corollary A: Suppose 1, 2, 3 are all executed, leaving 4 and 5. Then No. 4 gets the right to propose, he can put forward 100 of his own, and No. 5 does not give it.

Because of the rule restrictions, No. 5 is or against, and the vote between the two people is 1 person in favor and 1 person against, which is considered passed! So, when 4 and 5 are left, 4 will hold 100 and 5 will not.

Corollary B: Backward one level, when 1,2 are executed, and 3,4,5 are left, No. 3 can say the proposal that is most beneficial to himself and take care of No. 5 according to the above calculations - "99 for No. 3, No. 4 will not be given, and No. 5 will be 1". No. 5 had to agree because of the inference A.

2:1, the vote passed.

Corollary C: Backwards to the first level, when No. 1 is executed, leaving 2, 3, 4, and 5, No. 2 can say the proposal that is most beneficial to himself and take care of No. 4 according to the above calculations - "No. 2 99, No. 3 will not give, No. 4 will not give 1, No. 5 will not give". No. 4 had to agree because of the inference B.

3, 5 will naturally be against, but 2: 2, half passes.

Corollary D: No. 1's strategy - out of 5 people, No. 1 must get at least 3 votes, including himself. So number 1 can take care of number 3 and number 4.

Give No. 3 1, because if No. 3 disagrees, according to Corollary C, No. 3 will get nothing. Give No. 4 1, because if No. 4 disagrees, then either wait for No. 2 proposal and get 1 and get the same result as now; Or wait for Proposition 3 and get nothing.

So, the best proposal for No. 1 is: "98 on No. 1, 0 on No. 2, 1 on No. 3, 1 on No. 4, 0 on No. 5"!

1.If it's the first person, I want 33 diamonds, the first 3 people are divided, and the last 2 people can't help it!! Because they can't get through, they will be thrown into the sea to feed the sharks.

2。I'm the second person, I want to see how much the first person wants, improvisation!!

1。The first person wants 32 diamonds, 2 and 3 are 34 diamonds, the first 3 people are divided, and the last 2 people can't help it!!

2。I'm going to be the fourth person in front of all the people who vote and don't pass, if the others pass, there's no way, if they don't pass, hehe, all 100 votes, because it's already half.

No. 1 gets 56 gems, No. 2 and No. 3 don't have gems, and No. 4 and No. 5 each have 2 gems. Analysis: After the lottery decided the numbers, the attitude of the pirates was analyzed as follows:

Pirate 5 doesn't have a problem of saving his life, he just wants to get as many gems as possible. He wants to throw all four pirates into the sea to feed the sharks and take all the gems himself, but this is not possible. If it were Pirate 4's turn to propose a distribution plan, 5 would never agree to it, and 4 would definitely die at this point.

Therefore, No. 4 wanted to make a distribution plan before him. If it were Pirate 3's turn to come up with a distribution plan, he would definitely come up with a plan to take all the gems for himself. Because at this time, the 4th must vote yes to save his life.

Otherwise, as soon as No. 3 dies, No. 4 dies. Since it was Pirate 3's turn to propose a distribution plan, Pirates 4 and 5 would definitely not get a single gem, so they would find a way to pass the distribution proposed by 1 or 2. Of course, the premise is that you can get as many gems as possible.

If No. 1 died, No. 2 proposed that he must get 58 for himself, and give one to No. 4 and one to No. 5. In this way, No. 4 and No. 5 will agree, because once No. 2 dies, No. 3 will make a plan, and No. 4 and No. 5 will not accept the grains. No. 1 proposed that the distribution plan should be 92 for himself, and 4 for No. 4 and 4 for No. 5.

In this way, no matter what the attitude of No. 2 and No. 3 is, No. 4 and No. 5 will agree, because as soon as No. 1 dies, No. 2 can only give No. 4 and No. 5 2 each.

Pirate 1 gives 1 gold coin to No. 3, Hailstone No. 4 or 2 gold coins to No. 5, and he gets 97 gold coins for himself, i.e. (97,0,1,2,0) or (97,0,1,0,2). Now let's look at the rational analysis of each person as follows:

Start with Pirate 5 first, because he is the safest and does not risk being thrown into the sea, so his strategy is also the simplest, that is, it is best to have all the people in front of him die, then he can get the 100 gold coins alone.

Looking at No. 4 next, his chances of survival depend entirely on the survival of people in front of him, because if all the pirates from No. 1 to No. 3 feed the sharks, then with only No. 4 and No. 5 left, no matter what distribution plan No. 4 proposes, No. 5 will definitely vote against letting No. 4 feed the sharks in order to swallow all the gold coins. Even if 4 curries favor with 5 in order to save his life and proposes (0,100) to keep 5 alone for the gold, 5 may still feel that it is dangerous to keep 4 and vote against it to feed the sharks. Therefore, the rational No. 4 should not take such a risk and pin his hope of survival on the random selection of No. 5, and he can only guarantee his life by supporting No. 3.

Looking at No. 3 again, after the above logical reasoning, he will propose such a distribution plan as (100,0,0), because he knows that No. 4 will still unconditionally support him and vote in favor even if he gets nothing, so adding his own 1 vote can make him secure the 100 gold coins.

However, if No. 2 also learns about No. 3's allocation plan through reasoning, then he will propose a plan of (98,0,1,1). Because this plan is relative to the distribution plan of No. 3, No. 4 and No. 5 can get at least 1 gold coin, and the rational No. 4 and No. 5 will naturally feel that this plan is more beneficial to them and support No. 2, and do not want No. 2 to be out of the game and be distributed by No. 3. In this way, Number 2 can take 98 gold coins in a hurry.

Unfortunately, Pirate 1 is not a fuel-efficient lamp, and after some reasoning, he also has insight into the distribution plan of No. 2. The strategy he will adopt is to give up No. 2 and give No. 3 1 gold coin, and at the same time give No. 4 or No. 5 2 Jinling Heng coins, i.e., propose a distribution scheme of (97,0,1,2,0) or (97,0,1,0,2). Since the distribution plan of No. 1 can be more beneficial to No. 3 and No. 4 or No. 5 than the ruler plan of No. 2, then they will vote for No. 1, and with the vote of No. 1 itself, 97 gold coins can easily fall into No. 1's pocket.

10+1=11 (pcs).

80÷11=7……3 (pieces).

Therefore, A takes 3 pieces for the first time, and the sum of the number of pieces taken by B each time is 11, such as B takes 1 piece, A and B take 10 pieces, B takes 2 pieces, and A B takes 9 pieces

In order to make A win.

There should be 11 coins on the table when the last round B takes the coins.

For this reason, in the penultimate round, B should have 22 coins on the table.

For this reason, when the penultimate round B takes the coin, there should be 33 coins on the table.

For this reason, when the penultimate round B takes the coin, there should be 44 coins on the table.

For this reason, when the penultimate round B takes the coin, there should be 55 coins on the table.

For this reason, when the penultimate round B takes the coin, there should be 66 coins on the table.

For this reason, in the penultimate round, B should have 77 coins on the table when he takes the coin.

To do this, opening A should take out 3 coins.

- Analysis: 1) When there are 11 coins left on the table, B starts to take the coin, if B takes the minimum number 1, then A takes the maximum number of 10

If B takes the maximum number of 10, then A takes the minimum value of 1

2) When there are 22 coins left on the table, B starts to take the coins.

If B takes the minimum number 1, then A takes the maximum number 10 and the remainder is 11

If B takes the maximum number of 10, then A takes the minimum value of 1 and the remainder is 11

A takes 9 coins for the first time, and the rest is 10*x+1 anyway, which is 71, 61, 51, 41, 31, 21, 11, then you will definitely get the last one.

pieces are 2 3 of 16 17

The number of pieces of 2 3: 150 16 17 159 and the number of pieces of 3 8 divided by 2 3 is the original number of pieces.

159 and 3 8 2 3 159 and 3 8 3 2 239 and 1 16 A: The original 239 and 1 16 pieces.