微软经典面试题--海盗分宝石
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DickStrong
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Very interesting. I'm not a computer scientist, but my first reaction is:
Since 4 and 5 are safe, and after 1,2,3 all die, 4 will get all 100 diamonds, 4 will vote against any proposal from 1,2,3 that gives him less than 100 diamonds.
5 will get 0 if 1,2,3 all die, so he will vote for any proposal from 1,2,3 that gives him at least 1 diamonds.
That's the simplest, you geeks figure out the rest.
Since 4 and 5 are safe, and after 1,2,3 all die, 4 will get all 100 diamonds, 4 will vote against any proposal from 1,2,3 that gives him less than 100 diamonds.
5 will get 0 if 1,2,3 all die, so he will vote for any proposal from 1,2,3 that gives him at least 1 diamonds.
That's the simplest, you geeks figure out the rest.
DickStrong
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For 3, if 1,2 all die, he can simply give 5 one diamond to keep alive, and himself will get 99, so he will vote against any proposal from 1,2 that gives him less than 99.
For 2, if 1 dies, he can also keep himself alive by giving 5 one diamond, and himself will get 99, so he will also vote against proposal from 1 that gives him less than 99.
Now for 1, the only way to keep alive is to give 5 one diamond and 99 diamonds to either 2 or 3, so that he will get votes from 2/3 and 5 to keep alive.
Final results: 0-99-0-0-1 or 0-0-99-0-1.
For 2, if 1 dies, he can also keep himself alive by giving 5 one diamond, and himself will get 99, so he will also vote against proposal from 1 that gives him less than 99.
Now for 1, the only way to keep alive is to give 5 one diamond and 99 diamonds to either 2 or 3, so that he will get votes from 2/3 and 5 to keep alive.
Final results: 0-99-0-0-1 or 0-0-99-0-1.
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最初由 DickStrong 发布
Very interesting. I'm not a computer scientist, but my first reaction is:
Since 4 and 5 are safe, and after 1,2,3 all die, 4 will get all 100 diamonds, 4 will vote against any proposal from 1,2,3 that gives him less than 100 diamonds.
5 will get 0 if 1,2,3 all die, so he will vote for any proposal from 1,2,3 that gives him at least 1 diamonds.
That's the simplest, you geeks figure out the rest.
no
4 won't vote against any proposals. in fact 4 will vote for 2 if 2 give him one dimond
5 won't vote for anybody that gives him one dimond, in fact he will vote against 2 if 2 gives him one dimond
and you don't have to be a computer scientist to figure this out, you just need to pass grade 3
