两名球员正在玩剪刀石头布的一系列游戏。总共有N个游戏演奏的由阵列ARR [] []，其中ARR [I] [0]是播放机之一，ARR [I] [1]是玩家2的在第i个移动的移动表示集合{‘R’，’P’，’S’}中的游戏。任务是找到每个游戏的赢家。请注意，如果两个玩家都选择相同的项目，则该游戏为平局。
例子：
Input: arr[] = {“RS”, “SR”, “SP”, “PP”}
Output:
A
B
A
DRAW
Input: arr[] = {“SS”, “RP”, “PS”}
Output:
Draw
B
B
方法：假设播放器1由位1表示，播放器2由0表示。此外，让Rock表示为00 (十进制为0)，Paper表示为01 (十进制为1)，剪刀为10 (十进制为2)。
如果玩家选择岩石，它将以100表示，
同样， 101表示纸牌是由玩家一选择的。
前一位指示玩家编号，后两位指示其选择。
图案：
100(十进制4)(玩家1，摇滚)，001(十进制1)(玩家2，纸)->玩家2赢了(4-1 = 3)
101(十进制5)(玩家1，纸)，010(十进制2)(玩家2，剪刀)->玩家2赢了(5-2 = 3)
110(十进制6)(玩家1，剪刀)，000(十进制0)(玩家2，摇滚)->玩家2赢了(6-0 = 6)
101(十进制5)(玩家1，纸)，000(十进制0)(玩家2，摇滚)->玩家1赢了(5-0 = 5)
110(十进制6)(玩家1，剪刀)，001(十进制1)(玩家2，纸)->玩家1获胜(6-1 = 5)
100(十进制4)(玩家1，摇滚)，010(十进制2)(玩家2，剪刀)->玩家1赢了(4-2 = 2)
根据模式，如果差异是3的倍数，则玩家赢2次；如果差异为4，则游戏为平局。在其余情况下，玩家一会赢得比赛。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the
// winner of the game
string winner(string moves)
{
map data;
data['R'] = 0;
data['P'] = 1;
data['S'] = 2;
// Both the players chose to
// play the same move
if (moves[0] == moves[1]) {
return "Draw";
}
// Player A wins the game
if (((data[moves[0]] | 1 << (2))
- (data[moves[1]] | 0 << (2)))
% 3) {
return "A";
}
return "B";
}
// Function to perform the queries
void performQueries(string arr[], int n)
{
for (int i = 0; i < n; i++)
cout << winner(arr[i]) << endl;
}
// Driver code
int main()
{
string arr[] = { "RS", "SR", "SP", "PP" };
int n = sizeof(arr) / sizeof(string);
performQueries(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the
// winner of the game
static String winner(String moves)
{
HashMap data = new HashMap();
data.put('R', 0);
data.put('P', 1);
data.put('S', 2);
// Both the players chose to
// play the same move
if (moves.charAt(0) == moves.charAt(1))
{
return "Draw";
}
// Player A wins the game
if (((data.get(moves.charAt(0)) | 1 << (2)) -
(data.get(moves.charAt(1)) | 0 << (2))) % 3 != 0)
{
return "A";
}
return "B";
}
// Function to perform the queries
static void performQueries(String arr[], int n)
{
for (int i = 0; i < n; i++)
System.out.print(winner(arr[i]) + "\n");
}
// Driver code
public static void main(String[] args)
{
String arr[] = { "RS", "SR", "SP", "PP" };
int n = arr.length;
performQueries(arr, n);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 implementation of the approach
# Function to return the
# winner of the game
def winner(moves):
data = dict()
data['R'] = 0
data['P'] = 1
data['S'] = 2
# Both the players chose to
# play the same move
if (moves[0] == moves[1]):
return "Draw"
# Player A wins the game
if (((data[moves[0]] | 1 << (2)) -
(data[moves[1]] | 0 << (2))) % 3):
return "A"
return "B"
# Function to perform the queries
def performQueries(arr,n):
for i in range(n):
print(winner(arr[i]))
# Driver code
arr = ["RS", "SR", "SP", "PP"]
n = len(arr)
performQueries(arr, n)
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the
// winner of the game
static String winner(String moves)
{
Dictionary data = new Dictionary();
data.Add('R', 0);
data.Add('P', 1);
data.Add('S', 2);
// Both the players chose to
// play the same move
if (moves[0] == moves[1])
{
return "Draw";
}
// Player A wins the game
if (((data[moves[0]] | 1 << (2)) -
(data[moves[1]] | 0 << (2))) % 3 != 0)
{
return "A";
}
return "B";
}
// Function to perform the queries
static void performQueries(String []arr, int n)
{
for (int i = 0; i < n; i++)
Console.Write(winner(arr[i]) + "\n");
}
// Driver code
public static void Main(String[] args)
{
String []arr = { "RS", "SR", "SP", "PP" };
int n = arr.Length;
performQueries(arr, n);
}
}
// This code is contributed by 29AjayKumar A
B
A
Draw