通过添加数组中元素的两两差异找到赢家,直到可能
原文:https://www . geeksforgeeks . org/通过在可能的情况下添加成对的数组元素差来找到赢家/
给定一个正整数数组 arr[] ,两个玩家 A 和 B 在玩一个游戏。每次移动时,玩家从数组中选择两个数字 x 和 y ,如果数组中不存在| x–y |,则玩家将该数字添加到数组中(数组的大小增加 1)。不会移动的玩家输掉游戏。任务是如果玩家 A 总是开始游戏,找到游戏的赢家。 举例:
输入: arr[] = {2,3 } T3】输出:A A 移动后,阵将为{2,3,1},B 无法移动。 输入: arr[] = {5,6,7} 输出: B
进场:在这里观察到,在游戏结束时(当没有更多的移动可以进行时),结果数组将包含原始数组 gcd 的所有倍数,直到原始数组的最大元素。
例如,arr[] = {8,10} 自,gcd(8,10) = 2。所以游戏结束时得到的数组将包含所有 2 ≤ max(arr)的倍数,即 10。 因此,arr[] = {2,4,6,8,10}
从上面的观察,可以找到在原阵上可以执行的招式数,这将决定游戏的胜局,如果招式数为偶数那么 B 将是游戏的胜局否则 A 胜局。 移动次数可以发现为,(最大(arr)/gcd)–n其中 gcd 是原始数组元素的 gcd,最大(arr) / gcd 给出结果数组中元素的总数。从结果数组的元素计数中减去元素的原始计数将得到移动的次数。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the winner of the game
char getWinner(int arr[], int n)
{
// To store the gcd of the original array
int gcd = arr[0];
// To store the maximum element
// from the original array
int maxEle = arr[0];
for (int i = 1; i < n; i++) {
gcd = __gcd(gcd, arr[i]);
maxEle = max(maxEle, arr[i]);
}
int totalMoves = (maxEle / gcd) - n;
// If number of moves are odd
if (totalMoves % 2 == 1)
return 'A';
return 'B';
}
// Driver Code
int main()
{
int arr[] = { 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << getWinner(arr, n);
return 0;
}
Java 语言(一种计算机语言,尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to calculate gcd
static int __gcd(int a, int b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
// Function to return the winner
// of the game
static char getWinner(int []arr, int n)
{
// To store the gcd of the
// original array
int gcd = arr[0];
// To store the maximum element
// from the original array
int maxEle = arr[0];
for (int i = 1; i < n; i++)
{
gcd = __gcd(gcd, arr[i]);
maxEle = Math.max(maxEle, arr[i]);
}
int totalMoves = (maxEle / gcd) - n;
// If number of moves are odd
if (totalMoves % 2 == 1)
return 'A';
return 'B';
}
// Driver Code
public static void main(String args[])
{
int []arr = { 5, 6, 7 };
int n = arr.length;
System.out.print(getWinner(arr, n));
}
}
// This code is contributed
// by Akanksha Rai
Python 3
# Python3 implementation of the approach
from math import gcd
# Function to return the winner
# of the game
def getWinner(arr, n) :
# To store the gcd of the
# original array
__gcd = arr[0];
# To store the maximum element
# from the original array
maxEle = arr[0];
for i in range(1, n) :
__gcd = gcd(__gcd, arr[i]);
maxEle = max(maxEle, arr[i]);
totalMoves = (maxEle / __gcd) - n;
# If number of moves are odd
if (totalMoves % 2 == 1) :
return 'A';
return 'B';
# Driver Code
if __name__ == "__main__" :
arr = [ 5, 6, 7 ];
n = len(arr)
print(getWinner(arr, n))
# This code is contributed by Ryuga
C
// C# implementation of the approach
using System;
class GFG
{
// Function to calculate gcd
static int __gcd(int a, int b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
// Function to return the winner
// of the game
static char getWinner(int []arr, int n)
{
// To store the gcd of the
// original array
int gcd = arr[0];
// To store the maximum element
// from the original array
int maxEle = arr[0];
for (int i = 1; i < n; i++)
{
gcd = __gcd(gcd, arr[i]);
maxEle = Math.Max(maxEle, arr[i]);
}
int totalMoves = (maxEle / gcd) - n;
// If number of moves are odd
if (totalMoves % 2 == 1)
return 'A';
return 'B';
}
// Driver Code
public static void Main()
{
int []arr = { 5, 6, 7 };
int n = arr.Length;
Console.Write(getWinner(arr, n));
}
}
// This code is contributed
// by Akanksha Rai
服务器端编程语言(Professional Hypertext Preprocessor 的缩写)
<?php
// PHP implementation of the approach
// Function to calculate gcd
function __gcd($a, $b)
{
if ($b == 0)
return $a;
return __gcd($b, $a % $b);
}
// Function to return the winner
// of the game
function getWinner($arr, $n)
{
// To store the gcd of the
// original array
$gcd = $arr[0];
// To store the maximum element
// from the original array
$maxEle = $arr[0];
for ($i = 1; $i < $n; $i++)
{
$gcd = __gcd($gcd, $arr[$i]);
$maxEle = max($maxEle, $arr[$i]);
}
$totalMoves = ($maxEle / $gcd) - $n;
// If number of moves are odd
if ($totalMoves % 2 == 1)
return 'A';
return 'B';
}
// Driver Code
$arr = array(5, 6, 7);
$n = sizeof($arr);
echo getWinner($arr, $n);
// This code is contributed
// by Akanksha Rai
?>
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to calculate gcd
function __gcd(a, b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
// Function to return the winner
// of the game
function getWinner(arr, n)
{
// To store the gcd of the
// original array
let gcd = arr[0];
// To store the maximum element
// from the original array
let maxEle = arr[0];
for (let i = 1; i < n; i++)
{
gcd = __gcd(gcd, arr[i]);
maxEle = Math.max(maxEle, arr[i]);
}
let totalMoves = parseInt(maxEle / gcd, 10) - n;
// If number of moves are odd
if (totalMoves % 2 == 1)
return 'A';
return 'B';
}
let arr = [ 5, 6, 7 ];
let n = arr.length;
document.write(getWinner(arr, n));
</script>
Output:
B
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