通过给定的X和Y操作将N减少到0或更小

给定三个整数N ， X和Y，任务是检查是否可以通过以下操作将N减小到0或更小：

• 将N更新为⌊N/2⌋+ 10，最多X次
• 更新N到N – 10，最多y倍的情况。

例子：

Input: N = 100, X = 3, Y = 4
Output: Yes
Explanation:
Update N = 100 to ⌊100/2⌋ + 10 = 60.
Update N = 60 to 60 − 10 = 50.
Update N = 50 to ⌊50/2⌋ + 10 = 35.
Update N = 35 to ⌊35/2⌋ + 10 = 27.
Update N = 27 to 27 − 10 = 17. 
Update N = 17 to 17 − 10 = 7.
Update N = 7 to 7 − 10 = −3.

Input: N = 50, X = 1, Y = 2
Output: No

方法：可以根据以下观察结果解决此问题：

• 情况1：如果⌊N/2⌋+ 10> = N ，则仅执行第二次运算，因为第一次运算将增加值N。
• 情况2：如果先执行第一个操作，然后执行第二个操作，则结果为：

 
 

Operation 1: N = ⌊N/2⌋ + 10 
Operation 2: (⌊N/2⌋ + 10) – 10 = ⌊N/2⌋

 

• N的值减小为(⌊N/2⌋) 。
• 情况3：如果先执行第二项操作，然后执行第一项操作，则结果为：

 
 

Operation 2: N = N – 10 
Operation 1: ⌊(N – 10)/2⌋ + 10 = (⌊N/2⌋ – 5 + 10) = (⌊N/2⌋ + 5)

 

• N的值减小为(⌊N/2⌋+ 5) 。

从以上两个观察，如果N> N / 2 +10 ，则可以得出结论，为了将N的值减小到0或更小，必须在执行第二个操作以减小N的值之前执行第一个操作。
请按照以下步骤解决问题：

1. N的值更新为⌊N/2⌋+ 10如果N> N / 2 + 10和X> 0。
2. 更新的N到N的值– 10最多y倍的情况。
3. 检查N的更新值是否小于等于0。
4. 如果N≤0，则打印“是” 。否则，打印“否” 。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to check if N can be
// reduced to <= 0 or not
bool NegEqu(int N, int X, int Y)
{
 
    // Update N to N/2 + 10 at most X times
    while (X-- and N > N/2 + 10) {
        N = N / 2 + 10;
    }
 
    // Update N to N - 10 Y times
    while (Y--) {
        N = N - 10;
    }
 
    if (N <= 0)
        return true;
 
    return false;
}
 
// Driver Code
int main()
{
    int N = 100;
    int X = 3;
    int Y = 4;
 
    if (NegEqu(N, X, Y)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to check if N can be
// reduced to <= 0 or not
static boolean NegEqu(int N, int X, int Y)
{
     
    // Update N to N/2 + 10 at most X times
    while (X > 0 && (N > N / 2 + 10))
    {
        N = N / 2 + 10;
        X -= 1;
    }
 
    // Update N to N - 10 Y times
    while (Y > 0)
    {
        N = N - 10;
        Y -= 1;
    }
 
    if (N <= 0)
        return true;
 
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 100;
    int X = 3;
    int Y = 4;
 
    if (NegEqu(N, X, Y))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by jana_sayantan

Python3

# Python3 program to implement
# the above approach
 
# Function to check if N can be
# reduced to <= 0 or not
def NegEqu(N, X, Y):
 
    # Update N to N/2 + 10 at most X times
    while (X and N > N // 2 + 10):
        N = N // 2 + 10
        X -= 1
 
    # Update N to N - 10 Y times
    while (Y):
        N = N - 10
        Y -= 1
 
    if (N <= 0):
        return True
 
    return False
 
# Driver Code
if __name__ == '__main__':
     
    N = 100
    X = 3
    Y = 4
 
    if (NegEqu(N, X, Y)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to check if N can be
// reduced to <= 0 or not
public static bool NegEqu(int N, int X,
                          int Y)
{
     
    // Update N to N/2 + 10 at most X times
    while (X > 0 && (N > N / 2 + 10))
    {
        N = N / 2 + 10;
        X -= 1;
    }
 
    // Update N to N - 10 Y times
    while (Y > 0)
    {
        N = N - 10;
        Y -= 1;
    }
 
    if (N <= 0)
        return true;
 
    return false;
}
     
// Driver Code
public static void Main(String[] args)
{
    int N = 100;
    int X = 3;
    int Y = 4;
 
    if (NegEqu(N, X, Y))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by jana_sayantan

Javascript

 
// JavaScript implementation of the above approach
 
// Function to check if N can be
// reduced to <= 0 or not
function NegEqu(N, X, Y)
{
       
    // Update N to N/2 + 10 at most X times
    while (X > 0 && (N > N / 2 + 10))
    {
        N = N / 2 + 10;
        X -= 1;
    }
   
    // Update N to N - 10 Y times
    while (Y > 0)
    {
        N = N - 10;
        Y -= 1;
    }
   
    if (N <= 0)
        return true;
   
    return false;
}
 
// Driver code
         
    let N = 100;
    let X = 3;
    let Y = 4;
   
    if (NegEqu(N, X, Y))
    {
        document.write("Yes");
    }
    else
    {
        document.write("No");
    }
   
  // This code is contributed by code_hunt.

输出：

Yes

时间复杂度： O(X + Y)
辅助空间：O(1)