通过从中减去 D 或 1，检查 X 是否可以在精确的 T 次移动中减少到 0

给定一个整数X 、 D和T ，任务是检查是否有可能在 T 次移动中将 X 减少到 0。在每次移动中，X 可以减少 D 或 1。如果可能，打印 YES，否则打印 NO。

例子：

Input: X = 10, D = 3, T = 6
Output: YES
Explanation: Below are the moves applied- 
X = X – 3 = 7
X = X – 3 = 4
X = X – 1 = 3
X = X – 1 = 2
X = X – 1 = 1
X = X – 1 = 0
Hence it is possible to make X = 0 after exactly T moves.

Input: X = 10, D = 3, T = 5
Output: NO

朴素的方法：检查减少 D 和 1 的可能移动，并检查是否可以在 T 移动中使 X 变为 0。
时间复杂度： O(X)

有效的方法：给定的问题可以使用以下步骤解决：

• 设K为X减少D 的次数
• 所以总距离将是K*(D) + (T – K) ，它应该等于X 。
  因此方程变为

=> K*(D – 1) + T = X 
=> K*(D – 1) = X – T 
 

• 对于存在的有效答案，( X – T ) 应该可以被 ( D – 1 ) 整除

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 the above problem condition
int possibleReachingSequence(int X, int D, int T)
{
    // Check for base cases
    if (X < T) {
        cout << "NO";
        return 0;
    }
    if (T * D < X) {
        cout << "NO";
        return 0;
    }
    // Check if D - 1 is a divisor of X - T
    if ((X - T) % (D - 1) == 0) {
        cout << "YES";
    }
    else {
        cout << "NO";
    }
    return 0;
}
 
// Driver Code
int main()
{
    int X = 10, D = 3, T = 6;
    possibleReachingSequence(X, D, T);
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check the above problem condition
static int possibleReachingSequence(int X, int D, int T)
{
    // Check for base cases
    if (X < T) {
        System.out.println("NO");
        return 0;
    }
    if (T * D < X) {
        System.out.println("NO");
        return 0;
    }
    // Check if D - 1 is a divisor of X - T
    if ((X - T) % (D - 1) == 0) {
        System.out.println("YES");
    }
    else {
        System.out.println("NO");
    }
    return 0;
}
 
 
// Driver Code
public static void main(String args[])
{
    int X = 10, D = 3, T = 6;
    possibleReachingSequence(X, D, T);
}
}

Python3

# Python program for the above approach
 
# Function to check the above problem condition
def possibleReachingSequence(X, D, T):
   
  # Check the base case.
  if X < T:
    return "NO"
  if T*D < X:
    return "NO"
   
  # check if X-T is a divisor of D-1
  if (X - T) % (D - 1) == 0:
    return "YES"
  return "NO"
 
# Driver code
X = 10
D = 3
T = 6
print(possibleReachingSequence(X, D, T))
 
# This code is contributed by maddler.

C#

// C# program for the above approach
 
using System;
 
public class GFG{
 
// Function to check the above problem condition
static int possibleReachingSequence(int X, int D, int T)
{
    // Check for base cases
    if (X < T) {
        Console.WriteLine("NO");
        return 0;
    }
    if (T * D < X) {
        Console.WriteLine("NO");
        return 0;
    }
    // Check if D - 1 is a divisor of X - T
    if ((X - T) % (D - 1) == 0) {
        Console.WriteLine("YES");
    }
    else {
        Console.WriteLine("NO");
    }
    return 0;
}
 
 
// Driver Code
public static void Main(string []args)
{
    int X = 10, D = 3, T = 6;
    possibleReachingSequence(X, D, T);
}
}
 
// This code is contributed by AnkThon

输出

YES

时间复杂度： O(1)
辅助空间： O(1)