X1
发帖数: 1823 | 1 "You have a 100 coins laying flat on a table, each with a head side and a
tail side. 10 of them are heads up, 90 are tails up. You can't feel, see or
in any other way find out which side is up. Split the coins into two piles
such that there are the same number of heads in each pile."
是不是就是把每个coin切成两半，然后每个coin一边一半 |
l*******s
发帖数: 7316 | 2 如果题目沒少字，只要分成两堆各50个硬币即可。因为每个硬币都有头
："You have a 100 coins laying flat on a table, each with a head side
and a
：tail side. 10 of them are heads up, 90 are tails up. You can't feel,
see or in any other way find out which side is up. Split the coins into two
piles |
q*********o
发帖数: 1299 | 3 分两堆，10个一堆，90个一堆，再把其中一堆的面全部翻一下即可。
or
【在 X1 的大作中提到】
: "You have a 100 coins laying flat on a table, each with a head side and a
: tail side. 10 of them are heads up, 90 are tails up. You can't feel, see or
: in any other way find out which side is up. Split the coins into two piles
: such that there are the same number of heads in each pile."
: 是不是就是把每个coin切成两半，然后每个coin一边一半
|
z****g
发帖数: 3509 | 4 分两堆，90一堆，10一堆，再把10那一堆都翻一下。 |
q*********o
发帖数: 1299 | 5 刚改了一下
10个一堆，90个一堆
【在 X1 的大作中提到】
: "You have a 100 coins laying flat on a table, each with a head side and a
: tail side. 10 of them are heads up, 90 are tails up. You can't feel, see or
: in any other way find out which side is up. Split the coins into two piles
: such that there are the same number of heads in each pile."
: 是不是就是把每个coin切成两半，然后每个coin一边一半
|
X1
发帖数: 1823 | 6 这个正确
【在 z****g 的大作中提到】
: 分两堆，90一堆，10一堆，再把10那一堆都翻一下。
|
X1
发帖数: 1823 | 7 必须翻10个那堆
【在 q*********o 的大作中提到】
: 刚改了一下
: 10个一堆，90个一堆
|
q*********o
发帖数: 1299 | 8 不需要。哪对都可以。
【在 X1 的大作中提到】
: 必须翻10个那堆
|
X1
发帖数: 1823 | |
X1
发帖数: 1823 | 10 翻90那堆是不行的
8H2Tand 2H88T，一翻成了2T88H H就不一样了
【在 q*********o 的大作中提到】
: 不需要。哪对都可以。
|
z****g
发帖数: 3509 | 11 全都竖起来
【在 X1 的大作中提到】
: 如果不准翻面，那么是不是只能用刀切了？
|
X1
发帖数: 1823 | 12 good idea，但是不准翻面包括不准竖吧
【在 z****g 的大作中提到】
: 全都竖起来
|
l*******s
发帖数: 7316 | 13 分成10个和90的两堆，再把10个的全部翻一下
：分两堆，50个一堆，再把其中一堆的面全部翻一下。
： |
z******a
发帖数: 366 | 14 a_i = 1 if heads, 0 if tail
a_0 + a_1 + ... + a_n = 10
a_10 + a_11 + ... a_n = 10 - ( a_0 + a_1 + ... a_9 )
a_10 + a_11 + ... a_n = (1 - a_0) + (1 - a_1) + ... (1 - a_9)
flip any 10 coins will have the same number of heads of the rest. |
