反复从一堆中取出 2 枚硬币，从另一堆中取出 1 枚硬币，检查两堆硬币是否可以清空
原文:https://www . geeksforgeeks . org/check-if-两堆硬币-可以通过从一堆中反复取出 2 个硬币并从另一堆中取出 1 个硬币来清空/
给定两堆由硬币数量 N 和 M 组成的硬币，任务是通过重复从一堆中取出 2 枚硬币和从另一堆中取出 1 枚硬币来检查两堆硬币是否可以同时清空。如果两堆都可以清空，则打印“是”。否则，打印“否”。
示例:
输入: N = 1，M = 2 输出:是 说明: 从 1 st 堆中取出 1 枚硬币，从另一堆中取出 2 枚硬币。因此，两堆硬币的数量都是 0。
输入: N = 2，M = 2 T3】输出:否
方法:给定的问题可以基于以下观察来解决:
- 从堆中取出硬币有两种方式，即从堆 1 取出 2 枚硬币，从堆 2 取出 1 枚硬币，或者从堆 1 取出 1 枚硬币，从堆 2 取出 2 枚硬币。
- 设 X 为第一次移出的步数， Y 为第二次移出的步数，则两堆中剩余的硬币数分别为(N–2 * X–Y)和(M–2 * Y–X)。
- 执行 (X + Y) 移动后，必须清空堆，即:
(N–2 * X–Y)= 0 和(M–2 * Y–X)= 0
重新排列上述等式中的项会生成以下等式:
N = 2X + Y，M = 2Y + X
- 两堆硬币总数为(N+M)=(2 * X+Y)+(X+2 * Y)= 3(X+Y)。
- 因此，要使两个堆都为空，以下是条件:
因此，思路是检查硬币总数之和是否为3T3 的倍数，最大硬币数最多为最小硬币数的两倍，然后打印“是”。否则，打印“否”。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if two given piles
// can be emptied by repeatedly removing
// 2 coins from a pile and 1 coin from the other
void canBeEmptied(int A, int B)
{
// If maximum of A & B exceeds
// the twice of minimum of A & B
if (max(A, B) > 2 * min(A, B)) {
// Not possible to
// empty the piles
cout << "No";
return;
}
// If sum of both the coins is
// divisible by 3, then print Yes
if ((A + B) % 3 == 0)
cout << "Yes";
// Otherwise, print "No"
else
cout << "No";
}
// Driver Code
int main()
{
int A = 1, B = 2;
canBeEmptied(A, B);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for above approach
import java.util.*;
class GFG{
// Function to check if two given piles
// can be emptied by repeatedly removing
// 2 coins from a pile and 1 coin from the other
static void canBeEmptied(int A, int B)
{
// If maximum of A & B exceeds
// the twice of minimum of A & B
if (Math.max(A, B) > 2 * Math.min(A, B))
{
// Not possible to
// empty the piles
System.out.println("No");
return;
}
// If sum of both the coins is
// divisible by 3, then print Yes
if ((A + B) % 3 == 0)
System.out.println("Yes");
// Otherwise, print "No"
else
System.out.println("No");
}
// Driver Code
public static void main (String[] args)
{
int A = 1, B = 2;
canBeEmptied(A, B);
}
}
// This code is contributed by sanjoy_62
Python 3
# Python3 program for the above approach
# Function to check if two given piles
# can be emptied by repeatedly removing
# 2 coins from a pile and 1 coin from the other
def canBeEmptied(A, B):
# If maximum of A & B exceeds
# the twice of minimum of A & B
if (max(A, B) > 2 * min(A, B)):
# Not possible to
# empty the piles
print("No")
return
# If sum of both the coins is
# divisible by 3, then print Yes
if ((A + B) % 3 == 0):
print("Yes")
# Otherwise, print "No"
else:
print("No")
# Driver Code
if __name__ == '__main__':
A = 1
B = 2
canBeEmptied(A, B)
# This code is contributed by bgangwar59.
C
// C# program for the above approach
using System;
class GFG{
// Function to check if two given piles
// can be emptied by repeatedly removing
// 2 coins from a pile and 1 coin from the other
static void canBeEmptied(int A, int B)
{
// If maximum of A & B exceeds
// the twice of minimum of A & B
if (Math.Max(A, B) > 2 * Math.Min(A, B))
{
// Not possible to
// empty the piles
Console.WriteLine("No");
return;
}
// If sum of both the coins is
// divisible by 3, then print Yes
if ((A + B) % 3 == 0)
Console.WriteLine("Yes");
// Otherwise, print "No"
else
Console.WriteLine("No");
}
// Driver Code
public static void Main ()
{
int A = 1, B = 2;
canBeEmptied(A, B);
}
}
// This code is contributed by mohit kumar 29
java 描述语言
<script>
// Javascript program for the above approach
// Function to check if two given piles
// can be emptied by repeatedly removing
// 2 coins from a pile and 1 coin from the other
function canBeEmptied(A, B)
{
// If maximum of A & B exceeds
// the twice of minimum of A & B
if (Math.max(A, B) > 2 * Math.min(A, B))
{
// Not possible to
// empty the piles
document.write("No");
return;
}
// If sum of both the coins is
// divisible by 3, then print Yes
if ((A + B) % 3 == 0)
document.write("Yes");
// Otherwise, print "No"
else
document.write("No");
}
// Driver Code
let A = 1, B = 2;
canBeEmptied(A, B);
// This code is contributed by subhammahato348
</script>
Output:
Yes
时间复杂度:O(1) T5辅助空间:** O(1)
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