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There are n𝑛 coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If there is only one coin left, no coins will be flipped. If there are no facing-up coins, the player loses.Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.InputEach test contains multiple test cases. The first line contains the number of test cases t𝑡 (1≤t≤1001≤𝑡≤100). The description of the test cases follows.The first line of each test case contains only one positive integer n𝑛 (1≤n≤1001≤𝑛≤100), representing the number of the coins.A string s𝑠 of length n𝑛 follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.OutputFor each test case, print "YES" if Alice will win the game, and "NO" otherwise.You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.ExampleinputCopy35UUDUD5UDDUD2UUoutputCopyYESNONO

Question

There are n𝑛 coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If there is only one coin left, no coins will be flipped. If there are no facing-up coins, the player loses.Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.InputEach test contains multiple test cases. The first line contains the number of test cases t𝑡 (1≤t≤1001≤𝑡≤100). The description of the test cases follows.The first line of each test case contains only one positive integer n𝑛 (1≤n≤1001≤𝑛≤100), representing the number of the coins.A string s𝑠 of length n𝑛 follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.OutputFor each test case, print "YES" if Alice will win the game, and "NO" otherwise.You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.ExampleinputCopy35UUDUD5UDDUD2UUoutputCopyYESNONO

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Solution

The problem is a game theory problem involving two players, Alice and Bob, who take turns flipping coins arranged in a circle. The rules of the game are as follows:

  1. A player must choose a coin that is facing up, remove it, and then flip the two coins adjacent to it.
  2. If only two coins are left, one is removed and the other is not flipped (as it would be flipped twice).
  3. If only one coin is left, no coins are flipped.
  4. If there are no coins facing up, the player loses.

The goal is to determine who will win the game if both players play optimally. The game is guaranteed to end in a finite number of operations.

The input for the problem is a series of test cases. Each test case consists of a number representing the number of coins and a string representing the initial state of the coins ("U" for up, "D" for down).

The output for each test case is "YES" if Alice will win the game and "NO" otherwise.

Here's how you might approach solving this problem:

  1. For each test case, count the number of "U" in the string. This represents the number of coins that are initially facing up.
  2. If the number of "U" is odd, Alice will win. This is because Alice always goes first and can always ensure that she leaves an even number of "U" for Bob on each of her turns. Since Bob cannot change an even number of "U" to an odd number, Alice will always have a "U" to play on her turn.
  3. If the number of "U" is even, Bob will win. This is because no matter how Alice plays, she will always leave an odd number of "U" for Bob on her turn. Bob can then play optimally to ensure that he leaves an even number of "U" for Alice, eventually forcing Alice to face a situation with no "U" to play.
  4. Print "YES" if Alice will win and "NO" otherwise.

This solution assumes that both players play optimally and that Alice goes first.

This problem has been solved

Similar Questions

Alice, Bob, and Confucius are bored during recess, so they decide to play a new game. Each of them puts a dollar in the pot, and each tosses a coin. Alice wins if the coins land all heads or all tails. Bob wins if two heads and one tail land, and Confucius wins if one head and two tails land. The coins are fair, and the winner receives a net payment of $2 ($3 - $1 = $2), and the losers lose their $1.

Three balanced coins are flipped independently. One of the variables of interest is X, thenumber of heads. Let Y denote the amount of money won on a side bet in the followingmanner. If the first head occurs on the first flip, you win $1. If the first head occurs onthe second flip you win $2 and if the first head occurs on the third flip you win $3. Ifno heads appear, you lose $1 (that is, you win −$1).(a) [3 marks] In a table, list all possible outcomes of the experiment, along with thevalues of X and Y associated with each outcome.(b) [3 marks] Determine the bivariate distribution (that is, the joint probability dis-tribution) of X and Y . You can list the probabilities in a table.(c) [3 marks] Find the probability that fewer than three heads will occur and you willwin $1 or less.(d) [2 marks] Are X and Y independent? Why or why no

Yet Another Coin Problem

You have been offered to play a game. In this game, there are n𝑛 possible outcomes, and for each of them, you must bet a certain integer amount of coins. In the event that the i𝑖-th outcome turns out to be winning, you will receive back the amount of coins equal to your bet on that outcome, multiplied by ki𝑘𝑖. Note that exactly one of the n𝑛 outcomes will be winning.Your task is to determine how to distribute the coins in such a way that you will come out ahead in the event of any winning outcome. More formally, the total amount of coins you bet on all outcomes must be strictly less than the number of coins received back for each possible winning outcome.InputEach test consists of multiple test cases. The first line contains a single integer t𝑡 (1≤t≤1041≤𝑡≤104) — the number of test cases. The description of the test cases follows.The first line of each test case contains a single integer n𝑛 (1≤n≤501≤𝑛≤50) — the number of outcomes.The second line of each test case contains n𝑛 integers k1,k2,…,kn𝑘1,𝑘2,…,𝑘𝑛 (2≤ki≤202≤𝑘𝑖≤20) — the multiplier for the amount of coins if the i𝑖-th outcome turns out to be winning.It is guaranteed that the sum of n𝑛 over all test cases does not exceed 2⋅1052⋅105.OutputFor each test case, output −1−1 if there is no way to distribute the coins as required. Otherwise, output n𝑛 integers x1,x2,…,xn𝑥1,𝑥2,…,𝑥𝑛 (1≤xi≤1091≤𝑥𝑖≤109) — your bets on the outcomes.It can be shown that if a solution exists, there is always a solution that satisfies these constraints.If there are multiple suitable solutions, output any of them.ExampleinputCopy633 2 723 355 5 5 5 567 9 3 17 9 1336 3 259 4 6 8 3outputCopy27 41 12 1 1 -11989 1547 4641 819 1547 1071 -18 18 12 9 24NoteIn the first test case, the coins can be distributed as follows: 2727 coins on the first outcome, 4141 coins on the second outcome, 1212 coins on the third outcome. Then the total amount of coins bet on all outcomes is 27+41+12=8027+41+12=80 coins. If the first outcome turns out to be winning, you will receive back 3⋅27=813⋅27=81 coins, if the second outcome turns out to be winning, you will receive back 2⋅41=822⋅41=82 coins, if the third outcome turns out to be winning, you will receive back 7⋅12=847⋅12=84 coins. All these values are strictly greater than 8080.In the second test case, one way is to bet one coin on each of the outcomes.

In a game show called “Squad Game”, there are n contestants labelled 1 to n, in a circle.Moving clockwise around the circle, starting with contestant 1, each contestant k who has not yet been eliminated nominates a contestant (which may be themselves) for possible elimination. Contestant k then rolls a fair 6-sided die, and if the result is a 6, then the nominated player is eliminated from the game. Play continues in this way until a fixed number L < n of players have been eliminated.(a) Suppose that we watch this game and observe the labels of the L players eliminated in the order in which they were eliminated. Describe a suitable sample space for this experiment.(b) Suppose instead that we watch such a game and observe only the labels of the L players eliminated. Describe a suitable sample space for this experiment.(c) Suppose instead that we watch such a game and observe only whether player 1 is eliminated. Describe a suitable sample space for this experiment.(d) Suppose that each player nominates a uniformly chosen player among those who have not yet been eliminated (i.e. if j < L players have been eliminated then the player whose turn it currently rolls a fair (n − j)-sided die to determine who they will nominate). Find the probability of each sample point in the three experiments (a)-(c) above.

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