不使用ceil()函数查找a / b的ceil

在给定a和b的情况下，不使用上限函数即可找到a / b的上限值。

例子：

Input : a = 5, b = 4 
Output : 2 
Explanation: a/b = ceil(5/4) = 2 

Input : a = 10, b = 2
Output : 5 
Explanation: a/b = ceil(10/2) = 5 

使用ceiling函数可以解决该问题，但是当将整数作为参数传递时，ceiling函数不起作用。因此，下面有以下两种方法来找到上限值。

方法1：

ceilVal = (a / b) + ((a % b) != 0)

a / b返回整数除法值，并且(((a％b)！= 0)是一个检查条件，如果我们在a / b除法后还有余数，则返回1，否则返回0。整数除法值与检查值相加以获得最大值。

下面给出了上述方法的说明：

C++

// C++ program to find ceil(a/b)
// without using ceil() function
#include 
#include 
using namespace std;
  
// Driver function
int main()
{
    // taking input 1
    int a = 4;
    int b = 3;
    int val = (a / b) + ((a % b) != 0);
    cout << "The ceiling value of 4/3 is " 
         << val << endl;
  
    // example of perfect division
    // taking input 2
    a = 6;
    b = 3;
    val = (a / b) + ((a % b) != 0);
    cout << "The ceiling value of 6/3 is " 
         << val << endl;
  
    return 0;
}

Java

// Java program to find 
// ceil(a/b) without 
// using ceil() function
import java.io.*;
  
class GFG
{
    // Driver Code
    public static void main(String args[])
    {
        // taking input 1
        int a = 4;
        int b = 3, val = 0;
        if((a % b) != 0)
            val = (a / b) + 
                  (a % b);
        else
            val = (a / b);
        System.out.println("The ceiling " + 
                       "value of 4/3 is " + 
                                      val);
        // example of perfect
        // division taking input 2
        a = 6;
        b = 3;
        if((a % b) != 0)
            val = (a / b) + (a % b);
        else
            val = (a / b);
        System.out.println("The ceiling " + 
                       "value of 6/3 is " + 
                                      val);
    }
}
  
// This code is contributed by 
// Manish Shaw(manishshaw1)

Python3

# Python3 program to find ceil(a/b)
# without using ceil() function
import math
  
# Driver Code
  
# taking input 1
a = 4;
b = 3;
val = (a / b) + ((a % b) != 0);
print("The ceiling value of 4/3 is",
                   math.floor(val));
  
# example of perfect division
# taking input 2
a = 6;
b = 3;
val = int((a / b) + ((a % b) != 0));
print("The ceiling value of 6/3 is", val);
  
# This code is contributed by mits

C#

// C# program to find ceil(a/b)
// without using ceil() function
using System;
  
class GFG
{
  
    // Driver Code
    static void Main()
    {
        // taking input 1
        int a = 4;
        int b = 3, val = 0;
        if((a % b) != 0)
            val = (a / b) + (a % b);
        else
            val = (a / b);
        Console.WriteLine("The ceiling " + 
                      "value of 4/3 is " + 
                                     val);
      
        // example of perfect
        // division taking input 2
        a = 6;
        b = 3;
        if((a % b) != 0)
            val = (a / b) + (a % b);
        else
            val = (a / b);
        Console.WriteLine("The ceiling " + 
                      "value of 6/3 is " + 
                                     val);
    }
}
// This code is contributed by 
// Manish Shaw(manishshaw1)

输出：

The ceiling value of 4/3 is 2
The ceiling value of 6/3 is 2

方法二：

ceilVal = (a+b-1) / b

使用简单的数学运算，我们可以将分母加到分子上并从中减去1，然后将其除以分母得到上限值。

下面给出了上述方法的说明：

C++

// C++ program to find ceil(a/b)
// without using ceil() function
#include 
#include 
using namespace std;
  
// Driver function
int main()
{
    // taking input 1
    int a = 4;
    int b = 3;
    int val = (a + b - 1) / b;
    cout << "The ceiling value of 4/3 is " 
         << val << endl;
  
    // example of perfect division
    // taking input 2
    a = 6;
    b = 3;
    val = (a + b - 1) / b;
    cout << "The ceiling value of 6/3 is " 
         << val << endl;
  
    return 0;
}

Java

// Java program to find ceil(a/b)
// without using ceil() function
  
class GFG {
      
// Driver Code
public static void main(String args[])
{
      
    // taking input 1
    int a = 4;
    int b = 3;
    int val = (a + b - 1) / b;
    System.out.println("The ceiling value of 4/3 is "
                        + val);
  
    // example of perfect division
    // taking input 2
    a = 6;
    b = 3;
    val = (a + b - 1) / b;
    System.out.println("The ceiling value of 6/3 is "
                        + val );
}
}
  
// This code is contributed by Jaideep Pyne

Python3

# Python3 program to find 
# math.ceil(a/b) without 
# using math.ceil() function
import math
  
# Driver Code
# taking input 1
a = 4;
b = 3;
val = (a + b - 1) / b;
print("The ceiling value of 4/3 is ", 
                    math.floor(val));
  
# example of perfect division
# taking input 2
a = 6;
b = 3;
val = (a + b - 1) / b;
print("The ceiling value of 6/3 is ", 
                    math.floor(val));
  
# This code is contributed by mits

C＃

// C# program to find ceil(a/b)
// without using ceil() function
using System;
  
class GFG {
      
    // Driver Code
    public static void Main()
    {
          
        // taking input 1
        int a = 4;
        int b = 3;
        int val = (a + b - 1) / b;
        Console.WriteLine("The ceiling"
          + " value of 4/3 is " + val);
      
        // example of perfect division
        // taking input 2
        a = 6;
        b = 3;
        val = (a + b - 1) / b;
        Console.WriteLine("The ceiling"
         + " value of 6/3 is " + val );
    }
}
  
// This code is contributed by anuj_67.

输出：

The ceiling value of 4/3 is 2
The ceiling value of 6/3 is 2