将数字分成N个部分，以使最小和最大部分之间的差最小

给定两个整数“ X”和“ N”，任务是将整数“ X”分成正好为“ N”的部分，从而：
X1 + X2 + X3 + ... + Xn = X并且序列中最大和最小数字之间的差异最小。
如果不能将数字精确地划分为“ N”个部分，则最后打印该序列，然后打印“ -1”。

例子：

Input: X = 5, N = 3
Output: 1 2 2
Divide 5 into 3 parts such that the difference between the largest and smallest integer among
them is as minimal as possible. So we divide 5 as 1 + 2 + 2.

Input: X = 25, N = 5
Output: 5 5 5 5 5

方法：如果X >= N总是有一种分割数字的方法。

• 如果将数字精确地划分为“ N”个部分，则每个部分的值将为X/N ，剩余的X%N部分可以分配给任何X%N数字。
• 因此，如果X % N == 0则最小差将始终为’0’，并且该序列将包含所有相等的数字，即x/n 。
• 否则，差将为“ 1”，并且序列将为X/N, X/N, ..., (X/N)+1, (X/N)+1. 。

下面是上述方法的实现：

C++

// CPP implementation of the approach
#include
using namespace std;;
  
// Function that prints 
// the required sequence
void split(int x, int n)
{
  
// If we cannot split the 
// number into exactly 'N' parts
if(x < n)
cout<<"-1"<<" ";
  
          
  
    // If x % n == 0 then the minimum 
    // difference is 0 and all 
    // numbers are x / n
    else if (x % n == 0)
    {
        for(int i=0;i= zp)
            cout<<(pp + 1)<<" ";
            else
            cout<

---

Java

// Java implementation of the approach
   
class GFG{
// Function that prints 
// the required sequence
static void split(int x, int n)
{
   
// If we cannot split the 
// number into exactly 'N' parts
if(x < n)
System.out.print("-1 ");
   
           
   
    // If x % n == 0 then the minimum 
    // difference is 0 and all 
    // numbers are x / n
    else if (x % n == 0)
    {
        for(int i=0;i= zp)
            System.out.print((pp + 1)+" ");
            else
            System.out.print(pp+" ");
        }
    }
}
       
// Driver code
public static void main(String[] args)
{ 
           
int x = 5;
int n = 3;
split(x, n);
   
}
}
//This code is contributed by mits

Python3

# Python3 implementation of the approach
  
# Function that prints 
# the required sequence
def split(x, n):
  
    # If we cannot split the 
    # number into exactly 'N' parts
    if(x < n): 
        print(-1)
  
    # If x % n == 0 then the minimum 
    # difference is 0 and all 
    # numbers are x / n
    elif (x % n == 0):
        for i in range(n):
            print(x//n, end =" ")
    else:
        # upto n-(x % n) the values 
        # will be x / n 
        # after that the values 
        # will be x / n + 1
        zp = n - (x % n)
        pp = x//n
        for i in range(n):
            if(i>= zp):
                print(pp + 1, end =" ")
            else:
                print(pp, end =" ")
        
# Driver code          
x = 5
n = 3
split(x, n)

C#

// C# implementation of the approach
using System;
  
public class GFG{
    // Function that prints 
// the required sequence 
static void split(int x, int n) 
{ 
  
// If we cannot split the 
// number into exactly 'N' parts 
if(x < n) 
Console.WriteLine("-1 "); 
  
          
  
    // If x % n == 0 then the minimum 
    // difference is 0 and all 
    // numbers are x / n 
    else if (x % n == 0) 
    { 
        for(int i=0;i= zp) 
            Console.Write((pp + 1)+" "); 
            else
            Console.Write(pp+" "); 
        } 
    } 
} 
      
// Driver code 
static public void Main (){
  
int x = 5; 
int n = 3; 
split(x, n); 
  
} 
} 
//This code is contributed by Sachin.

PHP

= $zp)
            {
                echo (int)$pp + 1;
                echo (" ");
            }
            else
            {
                echo (int)$pp;
                echo (" ");
            }
        }
    }
}
  
// Driver code     
$x = 5;
$n = 3;
split( $x, $n);
  
// This code is contributed 
// by Shivi_Aggarwal
?>

输出：

1 2 2