两名球员正在玩剪刀石头布的一系列游戏。一共有K个游戏。玩家1具有以字符串A表示的字符串移动，而玩家2同样具有字符串B。如果任何玩家到达其字符串的末尾，他们将移回字符串的开头。任务是计算正好进行K场比赛时每个玩家赢得的比赛数。
例子：
Input: k = 4, a = “SR”, b = “R”
Output: 0 2
Game 1: Player1 = S, Player2 = R, Winner = Player2
Game 2: Player1 = R, Player2 = R, Winner = Draw
Game 3: Player1 = S, Player2 = R, Winner = Player2
Game 4: Player1 = R, Player2 = R, Winner = Draw
Input: k = 3, a = “S”, b = “SSS”
Output: 0 0
All the games are draws.
方法：将字符串a的长度设为n ，将字符串b的长度设为m 。这里的观察结果是，游戏将在n * m次移动后重复进行。因此，我们可以模拟n * m个游戏的过程，然后计算重复的次数。对于其余的游戏，由于它现在小于n * m ，因此我们可以再次模拟该过程。例如，在上面的第一个示例中， n = 2和m = 1 。因此，游戏将在每n * m = 2 * 1 = 2次移动后重复执行，即(Player2，Draw) ， (Player2，Draw) ，…..， (Player2，Draw) 。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
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 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 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
// 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 divyeshrabadiya07Python3
# 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 chitranayalC#
// 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 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 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 wins = countWins(k, a, b);
Console.WriteLine(wins.Item1 + " " +
wins.Item2);
}
}
// This code is contributed by divyesh072019 输出：
0 2时间复杂度： O(N * M)