一系列石头-纸-剪刀游戏中每个玩家的获胜次数
两名玩家正在玩石头-纸-剪刀系列游戏。一共玩了 K 游戏。玩家 1 有一系列由弦 A 表示的动作，类似地，玩家 2 有弦 B 。如果任何一个玩家到达了他们的绳子的末端，他们就会回到绳子的起点。任务是计算每个玩家在玩 K 游戏时的游戏数量。
示例:
输入: k = 4，a = "SR "，b = " R " T3】输出: 0 2 游戏 1:玩家 1 = S，玩家 2 = R，胜者=玩家 2 游戏 2:玩家 1 = R，玩家 2 = R，胜者=平局 游戏 3:玩家 1 = S，玩家 2 = R，胜者=玩家 2 游戏 4:玩家 1 = R，玩家 2 = R，胜者=平局
输入: k = 3，a =“S”，b =“SSS” T3】输出: 0 0 所有游戏均为平局。
进场:让弦长 a 为 n ，弦长 b 为 m 。这里的观察是游戏会在 n * m 移动后重复。所以，我们可以模拟 n * m 游戏的过程，然后统计它被重复的次数。对于剩下的游戏，我们可以再次模拟这个过程，因为它现在将小于 n * m 。例如，在上面的第一个例子中， n = 2 和 m = 1 。所以，游戏会在每一次 n * m = 2 * 1 = 2 移动后重复，即(玩家 2，抽奖)、(玩家 2，抽奖)、…..，(玩家 2，抽奖)。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns 1 if first player wins,
// 0 in case of a draw and -1 if second player wins
int compare(char first, char second)
{
// If both players have the same move
// then it's a draw
if (first == second)
return 0;
if (first == 'R') {
if (second == 'S')
return 1;
else
return -1;
}
if (first == 'P') {
if (second == 'R')
return 1;
else
return -1;
}
if (first == 'S') {
if (second == 'P')
return 1;
else
return -1;
}
}
// Function that returns the count of games
// won by both the players
pair<int, int> countWins(int k, string a, string b)
{
int n = a.length();
int m = b.length();
int i = 0, j = 0;
// Total distinct games that can be played
int moves = n * m;
pair<int, int> wins = { 0, 0 };
while (moves--) {
int res = compare(a[i], b[j]);
// Player 1 wins the current game
if (res == 1)
wins.first++;
// Player 2 wins the current game
if (res == -1)
wins.second++;
i = (i + 1) % n;
j = (j + 1) % m;
}
// Number of times the above n * m games repeat
int repeat = k / (n * m);
// Update the games won
wins.first *= repeat;
wins.second *= repeat;
// Remaining number of games after
// removing repeated games
int rem = k % (n * m);
while (rem--) {
int res = compare(a[i], b[j]);
// Player 1 wins the current game
if (res == 1)
wins.first++;
// Player 2 wins the current game
if (res == -1)
wins.second++;
i = (i + 1) % n;
j = (j + 1) % m;
}
return wins;
}
// Driver code
int main()
{
int k = 4;
string a = "SR", b = "R";
auto wins = countWins(k, a, b);
cout << wins.first << " " << wins.second;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.util.*;
import java.awt.Point;
class GFG{
// Function that returns 1 if first player wins,
// 0 in case of a draw and -1 if second player wins
public static int compare(char first, char second)
{
// If both players have the same move
// then it's a draw
if (first == second)
return 0;
if (first == 'R')
{
if (second == 'S')
return 1;
else
return -1;
}
if (first == 'P')
{
if (second == 'R')
return 1;
else
return -1;
}
if (first == 'S')
{
if (second == 'P')
return 1;
else
return -1;
}
return 0;
}
// Function that returns the count of games
// won by both the players
public static Point countWins(int k, String a,
String b)
{
int n = a.length();
int m = b.length();
int i = 0, j = 0;
// Total distinct games that
// can be played
int moves = n * m;
Point wins = new Point (0, 0);
while (moves-- > 0)
{
int res = compare(a.charAt(i),
b.charAt(j));
// Player 1 wins the current game
if (res == 1)
wins = new Point(wins.x + 1,
wins.y);
// Player 2 wins the current game
if (res == -1)
wins = new Point(wins.x,
wins.y + 1);
i = (i + 1) % n;
j = (j + 1) % m;
}
// Number of times the above
// n * m games repeat
int repeat = k / (n * m);
// Update the games won
wins = new Point(wins.x * repeat,
wins.y * repeat);
// Remaining number of games after
// removing repeated games
int rem = k % (n * m);
while (rem-- > 0)
{
int res = compare(a.charAt(i),
b.charAt(j));
// Player 1 wins the current game
if (res == 1)
wins = new Point(wins.x + 1,
wins.y);
// Player 2 wins the current game
if (res == -1)
wins = new Point(wins.x,
wins.y + 1);
i = (i + 1) % n;
j = (j + 1) % m;
}
return wins;
}
// Driver code
public static void main(String[] args)
{
int k = 4;
String a = "SR", b = "R";
Point wins = countWins(k, a, b);
System.out.println(wins.x + " " + wins.y);
}
}
// This code is contributed by divyeshrabadiya07
Python 3
# Python3 implementation of the above approach
# Function that returns 1 if first
# player wins, 0 in case of a draw
# and -1 if second player wins
def compare(first, second):
# If both players have the same
# move then it's a draw
if (first == second):
return 0
if (first == 'R'):
if (second == 'S'):
return 1
else:
return -1
if (first == 'P'):
if (second == 'R'):
return 1
else:
return -1
if (first == 'S'):
if (second == 'P'):
return 1
else:
return -1
# Function that returns the count
# of games won by both the players
def countWins(k, a, b):
n = len(a)
m = len(b)
i = 0
j = 0
# Total distinct games that
# can be played
moves = n * m
wins = [ 0, 0 ]
while (moves > 0):
res = compare(a[i], b[j])
# Player 1 wins the current game
if (res == 1):
wins[0] += 1
# Player 2 wins the current game
if (res == -1):
wins[1] += 1
i = (i + 1) % n
j = (j + 1) % m
moves -= 1
# Number of times the above
# n * m games repeat
repeat = k // (n * m)
# Update the games won
wins[0] *= repeat
wins[1] *= repeat
# Remaining number of games after
# removing repeated games
rem = k % (n * m)
while (rem > 0):
res = compare(a[i], b[j])
# Player 1 wins the current game
if (res == 1):
wins[0] += 1
# Player 2 wins the current game
if (res == -1):
wins[1] += 1
i = (i + 1) % n
j = (j + 1) % m
return wins
# Driver code
if __name__ == "__main__":
k = 4
a = "SR"
b = "R"
wins = countWins(k, a, b);
print(wins[0], wins[1])
# This code is contributed by chitranayal
C
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function that returns 1 if first player
// wins, 0 in case of a draw and -1 if
// second player wins
static int compare(char first, char second)
{
// If both players have the same
// move then it's a draw
if (first == second)
return 0;
if (first == 'R')
{
if (second == 'S')
return 1;
else
return -1;
}
if (first == 'P')
{
if (second == 'R')
return 1;
else
return -1;
}
if (first == 'S')
{
if (second == 'P')
return 1;
else
return -1;
}
return 0;
}
// Function that returns the count of games
// won by both the players
static Tuple<int, int> countWins(int k, string a,
string b)
{
int n = a.Length;
int m = b.Length;
int i = 0, j = 0;
// Total distinct games that
// can be played
int moves = n * m;
Tuple<int, int> wins = Tuple.Create(0, 0);
while (moves-- > 0)
{
int res = compare(a[i], b[j]);
// Player 1 wins the current game
if (res == 1)
wins = Tuple.Create(wins.Item1 + 1,
wins.Item2);
// Player 2 wins the current game
if (res == -1)
wins = Tuple.Create(wins.Item1,
wins.Item2 + 1);
i = (i + 1) % n;
j = (j + 1) % m;
}
// Number of times the above
// n * m games repeat
int repeat = k / (n * m);
// Update the games won
wins = Tuple.Create(wins.Item1 * repeat,
wins.Item2 * repeat);
// Remaining number of games after
// removing repeated games
int rem = k % (n * m);
while (rem-- > 0)
{
int res = compare(a[i], b[j]);
// Player 1 wins the current game
if (res == 1)
wins = Tuple.Create(wins.Item1 + 1,
wins.Item2);
// Player 2 wins the current game
if (res == -1)
wins = Tuple.Create(wins.Item1,
wins.Item2 + 1);
i = (i + 1) % n;
j = (j + 1) % m;
}
return wins;
}
// Driver Code
static void Main()
{
int k = 4;
string a = "SR", b = "R";
Tuple<int, int> wins = countWins(k, a, b);
Console.WriteLine(wins.Item1 + " " +
wins.Item2);
}
}
// This code is contributed by divyesh072019
java 描述语言
<script>
// Javascript implementation of the above approach
// Function that returns 1 if first player wins,
// 0 in case of a draw and -1 if second player wins
function compare(first,second)
{
// If both players have the same move
// then it's a draw
if (first == second)
return 0;
if (first == 'R')
{
if (second == 'S')
return 1;
else
return -1;
}
if (first == 'P')
{
if (second == 'R')
return 1;
else
return -1;
}
if (first == 'S')
{
if (second == 'P')
return 1;
else
return -1;
}
return 0;
}
// Function that returns the count of games
// won by both the players
function countWins(k,a,b)
{
let n = a.length;
let m = b.length;
let i = 0, j = 0;
// Total distinct games that
// can be played
let moves = n * m;
let wins = [0, 0];
while (moves-- > 0)
{
let res = compare(a[i],
b[j]);
// Player 1 wins the current game
if (res == 1)
wins = [wins[0] + 1,
wins[1]];
// Player 2 wins the current game
if (res == -1)
wins =[wins[0],
wins[1] + 1];
i = (i + 1) % n;
j = (j + 1) % m;
}
// Number of times the above
// n * m games repeat
let repeat = k / (n * m);
// Update the games won
wins = [wins[0] * repeat,
wins[1] * repeat];
// Remaining number of games after
// removing repeated games
let rem = k % (n * m);
while (rem-- > 0)
{
let res = compare(a[i],
b[j]);
// Player 1 wins the current game
if (res == 1)
wins = [wins[0] + 1,
wins[1]];
// Player 2 wins the current game
if (res == -1)
wins = [wins[0],
wins[1] + 1];
i = (i + 1) % n;
j = (j + 1) % m;
}
return wins;
}
// Driver code
let k = 4;
let a = "SR", b = "R";
let wins = countWins(k, a, b);
document.write(wins[0] + " " + wins[1]);
// This code is contributed by avanitrachhadiya2155
</script>
Output:
0 2
时间复杂度:O(N * M) T3】辅助空间: O(1)
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