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📜  游戏中最后一个怪物的最小可能最终HP Quality Center

📅  最后修改于: 2021-04-29 02:13:23             🧑  作者: Mango

给定N个怪兽，每个怪兽的初始生命值h [i]为整数。如果怪物的生命值大于0，则它仍然存在。在每一回合中，随机怪物杀死另一个随机怪物，即被攻击的怪物，其生命值会受到攻击怪物的生命值的影响而降低。这个过程一直持续到剩下一个怪物为止。最后剩下的怪物的最小可能生命值是多少。换句话说，任务是以这样的方式进行游戏：最终剩下的怪物的健康状况最低。

例子：

方法：从这个问题可以看出，一个人必须找到某种怪物的生命值，比方说，k可以杀死包括自我在内的其他怪物。一旦做出了至关重要的观察，问题就变得容易了。假设我们有两个怪兽，它们的生命值分别为h1和h2 ，例如h2> h1 。我们可以看到，在随机选择中，最佳方法是选择一个生命值较低的怪物，并降低另一个怪物的生命值，直到其生命值小于攻击怪物的生命值为止。此后，我们将选择第二只生命值小于h1的怪物，该过程将继续进行，直到只剩下一个怪物为止。因此，最后我们将剩下最小值gcd(h1，h2) 。这种gcd方法可以扩展到所有怪物。
因此，我们得出的怪物的最小可能健康状况将是给定怪物所有健康状况的gcd，即H(min)= gcd(h1，h2，…，hn)。

下面是上述方法的实现：

C++
// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the gcd of two numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to return the minimum
// possible health for the monster
int solve(int* health, int n)
{
    // gcd of first and second element
    int currentgcd = gcd(health[0], health[1]);
  
    // gcd for all subsequent elements
    for (int i = 2; i < n; ++i) {
        currentgcd = gcd(currentgcd, health[i]);
    }
    return currentgcd;
}
  
// Driver code
int main()
{
    int health[] = { 4, 6, 8, 12 };
    int n = sizeof(health) / sizeof(health[0]);
    cout << solve(health, n);
  
    return 0;
}

Java
// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function to return the gcd of two numbers
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to return the minimum
// possible health for the monster
static int solve(int health[], int n)
{
    // gcd of first and second element
    int currentgcd = gcd(health[0], health[1]);
  
    // gcd for all subsequent elements
    for (int i = 2; i < n; ++i) 
    {
        currentgcd = gcd(currentgcd, health[i]);
    }
    return currentgcd;
}
  
// Driver code
public static void main(String args[])
{
    int health[] = { 4, 6, 8, 12 };
    int n = health.length;
    System.out.println(solve(health, n));
}
}
  
// This code is contributed by
// Surendra_Gangwar

Python3
# Python3 implementation of the approach
  
# Function to return the gcd of two numbers
def gcd(a, b):
  
    if (a == 0):
        return b
    return gcd(b % a, a)
  
# Function to return the minimum
# possible health for the monster
def solve(health, n):
      
    # gcd of first and second element
    currentgcd = gcd(health[0], health[1])
  
    # gcd for all subsequent elements
    for i in range(2, n):
        currentgcd = gcd(currentgcd, 
                         health[i])
    return currentgcd
  
# Driver code
health = [4, 6, 8, 12]
n = len(health)
print(solve(health, n))
  
# This code is contributed by mohit kumar

C#
// C# implementation of the approach 
using System;
  
class GFG 
{ 
  
// Function to return the gcd of two numbers 
static int gcd(int a, int b) 
{ 
    if (a == 0) 
        return b; 
    return gcd(b % a, a); 
} 
  
// Function to return the minimum 
// possible health for the monster 
static int solve(int []health, int n) 
{ 
    // gcd of first and second element 
    int currentgcd = gcd(health[0], health[1]); 
  
    // gcd for all subsequent elements 
    for (int i = 2; i < n; ++i) 
    { 
        currentgcd = gcd(currentgcd, health[i]); 
    } 
    return currentgcd; 
} 
  
// Driver code 
public static void Main(String []args) 
{ 
    int []health = { 4, 6, 8, 12 }; 
    int n = health.Length; 
    Console.WriteLine(solve(health, n)); 
} 
} 
  
// This code is contributed by Arnab Kundu

PHP

输出： 
2