A. Bus Game

Codeforces

IDCF79A

Time2000ms

Memory256MB

Difficulty—

greedy

English · Original

Chinese · Translation

Formal · Original

After Fox Ciel won an onsite round of a programming contest, she took a bus to return to her castle. The fee of the bus was 220 yen. She met Rabbit Hanako in the bus. They decided to play the following game because they got bored in the bus. * Initially, there is a pile that contains _x_ 100-yen coins and _y_ 10-yen coins. * They take turns alternatively. Ciel takes the first turn. * In each turn, they must take exactly 220 yen from the pile. In Ciel's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 100-yen coins. In Hanako's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 10-yen coins. * If Ciel or Hanako can't take exactly 220 yen from the pile, she loses. Determine the winner of the game. ## Input The first line contains two integers _x_ (0 ≤ _x_ ≤ 106) and _y_ (0 ≤ _y_ ≤ 106), separated by a single space. ## Output If Ciel wins, print "_Ciel_". Otherwise, print "_Hanako_". [samples] ## Note In the first turn (Ciel's turn), she will choose 2 100-yen coins and 2 10-yen coins. In the second turn (Hanako's turn), she will choose 1 100-yen coin and 12 10-yen coins. In the third turn (Ciel's turn), she can't pay exactly 220 yen, so Ciel will lose.

在 Fox Ciel 赢得一场编程竞赛的现场赛段后，她乘坐巴士返回城堡。巴士票价为 220 日元。她在巴士上遇到了 Rabbit Hanako。由于在巴士上感到无聊，她们决定玩以下游戏。确定游戏的胜者。第一行包含两个整数 #cf_span[x] (#cf_span[0 ≤ x ≤ 106]) 和 #cf_span[y] (#cf_span[0 ≤ y ≤ 106])，用一个空格分隔。如果 Ciel 获胜，请输出 "_Ciel_"。否则，输出 "_Hanako_"。在第一轮（Ciel 的回合），她会选择 2 枚 100 日元硬币和 2 枚 10 日元硬币。在第二轮（Hanako 的回合），她会选择 1 枚 100 日元硬币和 12 枚 10 日元硬币。在第三轮（Ciel 的回合），她无法恰好支付 220 日元，因此 Ciel 输掉游戏。 ## Input 第一行包含两个整数 #cf_span[x] (#cf_span[0 ≤ x ≤ 106]) 和 #cf_span[y] (#cf_span[0 ≤ y ≤ 106])，用一个空格分隔。 ## Output 如果 Ciel 获胜，请输出 "_Ciel_"。否则，输出 "_Hanako_"。 [samples] ## Note 在第一轮（Ciel 的回合），她会选择 2 枚 100 日元硬币和 2 枚 10 日元硬币。在第二轮（Hanako 的回合），她会选择 1 枚 100 日元硬币和 12 枚 10 日元硬币。在第三轮（Ciel 的回合），她无法恰好支付 220 日元，因此 Ciel 输掉游戏。

**Definitions** Let $ x, y \in \mathbb{Z}_{\geq 0} $ denote the number of 100-yen and 10-yen coins available, respectively. Players alternate turns, starting with Ciel. On each turn, a player must select a non-negative integer number of 100-yen coins $ a $ and 10-yen coins $ b $ such that: $$ 100a + 10b = 220 $$ and the available coins satisfy $ a \leq \text{remaining 100-yen coins}, b \leq \text{remaining 10-yen coins} $. After the move, the used coins are removed from the pool. The player unable to make a valid move on their turn loses. **Constraints** $ 0 \leq x \leq 10^6 $, $ 0 \leq y \leq 10^6 $ **Objective** Determine the winner under optimal play: - If Ciel wins, output "Ciel". - If Hanako wins, output "Hanako". **Valid moves (solutions to $ 100a + 10b = 220 $)**: Simplify: $ 10a + b = 22 $, so $ b = 22 - 10a $, with $ a, b \in \mathbb{Z}_{\geq 0} $. Possible integer solutions: - $ a = 0 \Rightarrow b = 22 $ - $ a = 1 \Rightarrow b = 12 $ - $ a = 2 \Rightarrow b = 2 $ Thus, the set of valid moves is: $$ M = \{ (0,22), (1,12), (2,2) \} $$ The game is a two-player impartial game with state $ (x, y) $, and moves reduce $ x $ and $ y $ by amounts in $ M $. The player who cannot move loses.

Samples

Input #1

2 2

Output #1

Ciel

Input #2

3 22

Output #2

Hanako

API Response (JSON)

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Full JSON Raw Segments