通过移除最后一张给定的 N 张牌
找到赢得游戏的玩家
原文:https://www . geesforgeks . org/通过移除最后一张给定的 n 张牌来找到赢得游戏的玩家/
给定两个整数 N 和 K ，其中 N 表示游戏开始时出现的牌总数， K 表示单回合可以移除的最大牌数。两名玩家 A 和 B 轮流最多移除 K 张牌，从玩家 A 开始逐一移除，移除最后一张牌的玩家即为赢家。任务是检查 A 能否赢得比赛。如果发现是真的，打印‘A’作为答案。否则，打印‘B’。
示例:
输入: N = 14，K = 10 输出:是 说明: 回合 1: A 第一回合移除 3 张牌。 第 2 回合:B 从范围【1–10】 中移除任意数量的牌，最后 A 可以移除所有剩余牌并赢得游戏，因为第 2 回合后剩余牌的数量将≤ 10
输入: N = 11，K=10 输出:否
进场:这里的思路是观察每当的值为 N % (K + 1) = 0 时，那么 A 将永远无法赢得比赛。否则 A 永远赢不了比赛。 证明:
- If N ≤ K: The person who has the first round will win the game, that is, A. 。
- If n = k+1: A can remove any number of cards within the range of [1, k] . So the total number of cards left after the first round is also in the range of [1,k] . B now gets the turn, and the number of cards left is within the range of [1, k] . So, B will win the game.
- If K+2 ≤ N ≤ 2K+1: A removes N -( K+1) cards in the first round. B Any number of cards within the range of [1,k] can be removed in the next round. So now the total number of cards left is within the range of [1,k] . Now, because the remaining cards are in the range of [1, k] , A can remove all the cards and win the game.
所以思路是检查 N % (K + 1) 是否等于 0。如果发现是真的，打印 B 作为获胜者。否则，打印 A 作为获胜者。 以下是上述办法的实施情况:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check which
// player can win the game
void checkWinner(int N, int K)
{
if (N % (K + 1)) {
cout << "A";
}
else {
cout << "B";
}
}
// Driver code
int main()
{
int N = 50;
int K = 10;
checkWinner(N, K);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to check which
// player can win the game
static void checkWinner(int N, int K)
{
if (N % (K + 1) > 0)
{
System.out.print("A");
}
else
{
System.out.print("B");
}
}
// Driver code
public static void main(String[] args)
{
int N = 50;
int K = 10;
checkWinner(N, K);
}
}
// This code is contributed by Amit Katiyar
Python 3
# Python3 program to implement
# the above approach
# Function to check which
# player can win the game
def checkWinner(N, K):
if(N % (K + 1)):
print("A")
else:
print("B")
# Driver Code
N = 50
K = 10
# Function call
checkWinner(N, K)
# This code is contributed by Shivam Singh
C
// C# program to implement
// the above approach
using System;
class GFG{
// Function to check which
// player can win the game
static void checkWinner(int N, int K)
{
if (N % (K + 1) > 0)
{
Console.Write("A");
}
else
{
Console.Write("B");
}
}
// Driver code
public static void Main(String[] args)
{
int N = 50;
int K = 10;
checkWinner(N, K);
}
}
// This code is contributed by Amit Katiyar
java 描述语言
<script>
// Javascript Program to implement
// the above approach
// Function to check which
// player can win the game
function checkWinner(N, K)
{
if (N % (K + 1)) {
document.write("A");
}
else {
document.write("B");
}
}
// Driver code
let N = 50;
let K = 10;
checkWinner(N, K);
// This code is contributed by Saurabh Jaiswal
</script>
Output:
A
时间复杂度:O(1) T5辅助空间:** O(1)
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