I***e
发帖数: 1136 | 1 Play A and B play the following game: there are 100 random numbers in a row.
Player A chooses 1 from an end (that is, the first or the 100th). Then player
B choose from an end of the remaining 99 numbers. Then A does the same, and on
and on.
The who gets a bigger sum wins the game.
What's A's sure-win strategy?
-iCare- | m****n
发帖数: 45 | 2 Suppose there are 2n numbers x_1,...,x_{2n}.
If x_1+x_3+...+x_{2n-1}>= x_2+x_4+...+x_{2n}
Player A should choose x_1.
Because after you choose x_1, B can only choose a number with even subscript.
After B choose a number, you can choose a number with odd subscript again and
B can only choose a number with even subscript.
A keeps choosing numbers with odd subscript and B can only choose numbers with
even subscript.
At the end, A would not lose.
If x_1+...+x_{2n-1} < x_2+...+x_{2n}, Player A should
【在 I***e 的大作中提到】
: Play A and B play the following game: there are 100 random numbers in a row.
: Player A chooses 1 from an end (that is, the first or the 100th). Then player
: B choose from an end of the remaining 99 numbers. Then A does the same, and on
: and on.
: The who gets a bigger sum wins the game.
: What's A's sure-win strategy?
: -iCare-
| I***e
发帖数: 1136 | 3 Great!
.
and
with
【在 m****n 的大作中提到】
: Suppose there are 2n numbers x_1,...,x_{2n}.
: If x_1+x_3+...+x_{2n-1}>= x_2+x_4+...+x_{2n}
: Player A should choose x_1.
: Because after you choose x_1, B can only choose a number with even subscript.
: After B choose a number, you can choose a number with odd subscript again and
: B can only choose a number with even subscript.
: A keeps choosing numbers with odd subscript and B can only choose numbers with
: even subscript.
: At the end, A would not lose.
: If x_1+...+x_{2n-1} < x_2+...+x_{2n}, Player A should
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