在第i步中找到捐赠i糖果游戏的赢家

鉴于两个整数X和Y分别代表分配给玩家A和B的糖果，其中两个玩家在捐赠我的沉迷游戏的数量糖果对手在每一个^第i^个举动。从玩家A开始，游戏交替进行，直到玩家无法捐赠所需数量的糖果并输掉游戏为止，任务是找到游戏的获胜者。

例子：

Input: X = 2, Y = 3
Output: A
Explanation: The game turns out in the following manner:

Step	X	Y
0	2	3
1	1	4
2	3	2
3	0	5
4	4	1

Since A fails to donate 5 candies in the 5^th step, B wins the game.

 Input: X = 2, Y = 1
Output: B
Explanation: The game turns out in the following manner:

Step	X	Y
0	2	1
1	1	2
2	3	0
3	0	3

Since B fails to give 4 candies in the 4^th step, A wins the game.

方法：想法是基于以下观察结果解决问题：

Step	Number of candies Possessed by A	Number of candies Possessed by B
0	X	Y
1	X – 1	Y + 1
2	X – 1 + 2 = X + 1	Y + 1 – 2 = Y – 1
3	X – 2	Y + 2
4	X + 2	Y – 2

• The player whose candies reduce to 0 first, will not be having enough candies to give in the next move.
• Count of candies of player A decreases by 1 in odd moves.
• Count of candies of player B decreases by 1 in even moves.

请按照以下步骤解决问题：

1. 初始化两个变量，分别是机会A和机会B，代表玩家拥有的糖果数量减少到0的移动次数。
2. 由于玩家A的糖果数在奇数步中减少1，因此机会A = 2 *(X – 1)+ 1
3. 由于玩家B的糖果数在均匀移动中减少1，因此偶然机会B = 2 * Y
4. 如果机会A <机会B ，则B将是获胜玩家。
5. 否则， A将是获胜者。
6. 打印获胜的玩家。

下面是上述方法的实现：

C++

// C++ Program for the
// above approach
  
#include 
using namespace std;
  
// Function to find the winning
// player in a game of donating i
// candies to opponent in i-th move
int stepscount(int a, int b)
{
    // Steps in which number of
    // candies of player A finishes
    int chanceA = 2 * a - 1;
  
    // Steps in which number of
    // candies of player B finishes
    int chanceB = 2 * b;
  
    // If A's candies finishes first
    if (chanceA < chanceB) {
        cout << "B";
    }
  
    // Otherwise
    else if (chanceB < chanceA) {
        cout << "B";
    }
  
    return 0;
}
  
// Driver Code
int main()
{
    // Input
  
    // Candies possessed
    // by player A
    int A = 2;
  
    // Candies possessed
    // by player B
    int B = 3;
  
    stepscount(A, B);
  
    return 0;
}

输出：

B

时间复杂度： O(1)
辅助空间： O(1)