一到英文逻辑题two players play a game.we have coins whose value is 1,5,10,25,50,100.And the two players take turns to put one coin into a pot.The one who reach value 678 will be the winner (it is not allowed to exceed this value).who will win?

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一到英文逻辑题twoplayersplayagame.wehavecoinswhosevalueis1,5,10,25,50,100.Andthetwoplayerstaketurnstoputone

一到英文逻辑题two players play a game.we have coins whose value is 1,5,10,25,50,100.And the two players take turns to put one coin into a pot.The one who reach value 678 will be the winner (it is not allowed to exceed this value).who will win?
一到英文逻辑题
two players play a game.we have coins whose value is 1,5,10,25,50,100.And the two players take turns to put one coin into a pot.The one who reach value 678 will be the winner (it is not allowed to exceed this value).who will win?

一到英文逻辑题two players play a game.we have coins whose value is 1,5,10,25,50,100.And the two players take turns to put one coin into a pot.The one who reach value 678 will be the winner (it is not allowed to exceed this value).who will win?
两个玩家玩一种游戏.我们有面值为1,5,10,25,50,100的硬币.并且两个玩家依次轮流将一枚硬币投入一个罐子里.谁先投到678的总值谁就是赢家(不许超过678).谁将赢?
答:the second one
第二个投的有机会赢,也可能都赢不了(超过规定总值了)
只用看个位数就够了,只要能将个位数投到8的人就是胜利者
5+1+1+1=8
1+5+1+1=8
1+1+5+1=8
1+1+1+5=8
1+1+1+1+1+1+1+1=8
等等
所以是偶数号的玩家(也就是后投的玩家)有机会赢(只是时间问题而已),儿奇数号玩家(也就是先投的玩家)一定赢不了

二人做游戏,有面值1, 5, 10, 25, 50, 100的硬币,二人轮流投一个硬币入钵,谁先够678谁就是获胜者(不允许超过这个值).请问谁会赢?
我认为都是5的倍数面值,就首先投3次1,然后剩下的675用6次100,1次50,1次25.这样就是一共最少11次,先投够11次的人会赢.哈哈仅做参考!:)...

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二人做游戏,有面值1, 5, 10, 25, 50, 100的硬币,二人轮流投一个硬币入钵,谁先够678谁就是获胜者(不允许超过这个值).请问谁会赢?
我认为都是5的倍数面值,就首先投3次1,然后剩下的675用6次100,1次50,1次25.这样就是一共最少11次,先投够11次的人会赢.哈哈仅做参考!:)

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2个人玩钱币的游戏。钱币的面值是1,5,10,25,50,100. 2人轮流拿钱币放到瓮里面去,每次只能拿1个,计算总值, 谁刚好拿到 678(且不能超过678), 谁就获胜了。
也就是说,你只要拿到几个数,你就稳赢了,求那几个数值。
另一种想法就是让对方拿那几个数字,你稳赢的策略。
思想如此,太晚了。没有深入去考虑。。。
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2个人玩钱币的游戏。钱币的面值是1,5,10,25,50,100. 2人轮流拿钱币放到瓮里面去,每次只能拿1个,计算总值, 谁刚好拿到 678(且不能超过678), 谁就获胜了。
也就是说,你只要拿到几个数,你就稳赢了,求那几个数值。
另一种想法就是让对方拿那几个数字,你稳赢的策略。
思想如此,太晚了。没有深入去考虑。。。
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按耐不住,来补充下吧。
2次奇数的和才为偶数。也就是偶数次奇数的和才有可能是8.所以看谁出第1个奇数了,谁出了第1个奇数,谁就有输了。后出的获胜面很大。第1人不出奇数,第2人只要保证自己拿到的是偶数,就基本赢定了。而不是从怎样加起来是8这个出发点来考虑.
如果末尾是奇数,那么先手的人基本上赢定了。
另外就是要考虑硬币的面值。如果硬币面值全是偶数,就只有抓关键数了。
应该可以写篇论文。

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