表示为五个连续整数之和的数字

给定一个整数n，任务是确定n是否可以表示为五个连续整数的和。如果是，则找到五个连续的整数，否则打印“ -1”。
例子：

Input : n = 15
Output : 1 2 3 4 5
15 = 1 + 2 + 3 + 4 + 5

Input : n = 18
Output : -1

方法1 ：(强力)
想法是运行从i = 0到n – 4的循环，检查(i + i + 1 + i + 2 + i + 3 + i + 4)是否等于n。另外，请检查n是否为正数或负数，并相应地使i增加或减少1。
以下是此方法的实现：

C++

// CPP Program to check if a number can
// be expressed as sum of five consecutive
// integers.
#include 
using namespace std;
 
// function to check if a number can be expressed as
// sum of five consecutive integers.
void checksum(int n)
{
    // if n is 0
    if (n == 0) {
        cout << "-2 -1 0 1 2" << endl;
        return;
    }
 
    int inc;
 
    // if n is positive, increment loop by 1.
    if (n > 0)
        inc = 1;
 
    // if n is negative, decrement loop by 1.
    else
        inc = -1;
 
    // Running loop from 0 to n - 4
    for (int i = 0; i <= n - 4; i += inc) {
 
        // check if sum of five consecutive
        // integer is equal to n.
        if (i + i + 1 + i + 2 + i + 3 + i + 4 == n) {
            cout << i << " " << i + 1
                 << " " << i + 2
                 << " " << i + 3
                 << " " << i + 4;
            return;
        }
    }
 
    cout << "-1";
}
 
// Driver Program
int main()
{
    int n = 15;
    checksum(n);
    return 0;
}

Java

// Java Program to check if a number can
// be expressed as sum of five consecutive
// integers.
import java.io.*;
 
class GFG {
 
    // function to check if a number can be
    // expressed as sum of five consecutive
    // integers.
    static void checksum(int n)
    {
        // if n is 0
        if (n == 0) {
            System.out.println("-2 -1 0 1 2");
            return;
        }
     
        int inc;
     
        // if n is positive, increment loop by 1.
        if (n > 0)
            inc = 1;
     
        // if n is negative, decrement loop by 1.
        else
            inc = -1;
     
        // Running loop from 0 to n - 4
        for (int i = 0; i <= n - 4; i += inc) {
     
            // check if sum of five consecutive
            // integer is equal to n.
            if (i + i + 1 + i + 2 + i + 3 + i
                                       + 4 == n)
            {
                System.out.print( (i )
                        + " " + (i + 1)
                        + " " + (i + 2)
                        + " " + (i + 3)
                        + " " + (i + 4));
                return;
            }
        }
     
        System.out.println( "-1");
    }
     
    // Driver Program
    public static void main (String[] args)
    {
        int n = 15;
        checksum(n);
    }
}
 
// This code is contributed by anuj_67

Python3

# Python3 code to check if a number
# can be expressed as sum of five
# consecutive integers.
  
# function to check if a number
# can be expressed as sum of five
# consecutive integer.
def checksum(n):
  
    # if n is 0
    if n == 0:
        print("-2 -1 0 1 2")
        return 0
          
    inc = 0
  
    # if n is positive,
    # increment loop by 1.
    if n > 0:
        inc = 1
      
    # if n is negative,
    # decrement loop by 1.
    else:
        inc = -1
      
    # Running loop from 0 to n - 4
    for i in range(0, n-3, inc):
      
        # check if sum of five consecutive
        # integer is equal to n.
        if i + i + 1 + i + 2 + i + 3 + i + 4 == n:
            print(i, " ", i + 1, " ", i + 2, " ", i + 3, " ", i + 4)
            return 0
              
    print("-1")
      
# Driver Code
n = 15
checksum(n)

C#

// C# Program to check if a number can
// be expressed as sum of five consecutive
// integers.
using System;
 
class GFG {
 
    // function to check if a number can be
    // expressed as sum of five consecutive
    // integers.
    static void checksum(int n)
    {
        // if n is 0
        if (n == 0) {
            Console.Write("-2 -1 0 1 2");
            return;
        }
     
        int inc;
     
        // if n is positive, increment loop by 1.
        if (n > 0)
            inc = 1;
     
        // if n is negative, decrement loop by 1.
        else
            inc = -1;
     
        // Running loop from 0 to n - 4
        for (int i = 0; i <= n - 4; i += inc) {
     
            // check if sum of five consecutive
            // integer is equal to n.
            if (i + i + 1 + i + 2 + i + 3 + i
                                    + 4 == n)
            {
                Console.Write( (i )
                        + " " + (i + 1)
                        + " " + (i + 2)
                        + " " + (i + 3)
                        + " " + (i + 4));
                return;
            }
        }
     
        Console.WriteLine( "-1");
    }
     
    // Driver Program
    public static void Main ()
    {
        int n = 15;
        checksum(n);
    }
}
 
// This code is contributed by anuj_67

PHP

 0)
        $inc = 1;
 
    // if n is negative,
    // decrement loop by 1.
    else
        $inc = -1;
 
    // Running loop from
    // 0 to n - 4
    for ($i = 0;
         $i <= $n - 4; $i += $inc)
    {
 
        // check if sum of five
        // consecutive integer
        // is equal to n.
        if ($i + $i + 1 + $i + 2 +
            $i + 3 + $i + 4 == $n)
        {
            echo $i , " " , $i + 1,
                      " " , $i + 2,
                      " " , $i + 3,
                      " " , $i + 4;
            return;
        }
    }
 
    echo "-1";
}
 
// Driver Code
$n = 15;
checksum($n);
 
// This code is contributed
// by ajit
?>

Javascript

 
// JavaScript program to check if a number can
// be expressed as sum of five consecutive
// integers.
 
    // function to check if a number can be
    // expressed as sum of five consecutive
    // integers.
    function checksum(n) {
        // if n is 0
        if (n == 0) {
            document.write("-2 -1 0 1 2");
            return;
        }
 
        var inc;
 
        // if n is positive, increment loop by 1.
        if (n > 0)
            inc = 1;
 
        // if n is negative, decrement loop by 1.
        else
            inc = -1;
 
        // Running loop from 0 to n - 4
        for (i = 0; i <= n - 4; i += inc) {
 
            // check if sum of five consecutive
            // integer is equal to n.
            if (i + i + 1 + i + 2 + i + 3 + i + 4 == n) {
                document.write((i) + " "
                + (i + 1) + " "
                + (i + 2) + " "
                + (i + 3) + " "
                + (i + 4));
                return;
            }
        }
 
        document.write("-1");
    }
 
    // Driver Program
     
        var n = 15;
        checksum(n);
 
// This code contributed by gauravrajput1
 

输出 ：

1 2 3 4 5

方法2 ：(有效方法)
想法是检查n是否为5的倍数。
令n为k – 2，k-1，k，k + 1，k + 2的五个连续整数之和。所以，
k-2 + k-1 + k + k + 1 + k + 2 = n
5 * k = n
这五个数字将为n / 5 – 2，n / 5 – 1，n / 5，n / 5 + 1，n / 5 + 2。以下是此方法的实现：

C++

// CPP Program to check if a number can be
// expressed as sum of five consecutive integer.
#include 
using namespace std;
 
// function to check if a number can be
// expressed as sum of five consecutive
// integers.
void checksum(int n)
{
    // if n is multiple of 5
    if (n % 5 == 0)
        cout << n / 5 - 2 << " "
             << n / 5 - 1 << " " << n / 5
             << " " << n / 5 + 1 << " "
             << n / 5 + 2;
 
    // else print "-1".
    else
        cout << "-1";
}
 
// Driver Program
int main()
{
    int n = 15;
    checksum(n);
    return 0;
}

Java

// Java Program to check if a number can
// be expressed as sum of five consecutive
// integer.
import java.io.*;
 
class GFG {
 
    // function to check if a number can
    // be expressed as sum of five
    // consecutive integers.
    static void checksum(int n)
    {
        // if n is multiple of 5
        if (n % 5 == 0)
            System.out.println( (n / 5 - 2)
                  + " " + (n / 5 - 1) + " "
                  + (n / 5) + " " + (n / 5
                + 1 ) + " " + (n / 5 + 2));
     
        // else print "-1".
        else
            System.out.println( "-1");
    }
     
    // Driver Program
    public static void main (String[] args)
    {
        int n = 15;
        checksum(n);
    }
}
 
// This code is contributed by vt_m.

Python3

# Python3 code to check if a number
# can be expressed as sum of five
# consecutive integer.
  
# function to check if a number
# can be expressed as sum of five
# consecutive integers.
def checksum(n):
    n = int(n)
      
    # if n is multiple of 5
    if n % 5 == 0:
        print(int(n / 5 - 2), " ",
         int(n / 5 - 1), " ", int(n / 5), " ", int(n / 5 + 1), " ", int(n / 5 + 2))
      
    # else print "-1".
    else:
        print("-1")
          
# Driver Code
n = 15
checksum(n)

C＃

// C# Program to check if a number can
// be expressed as sum of five consecutive
// integer.
using System;
 
class GFG {
 
    // function to check if a number can
    // be expressed as sum of five
    // consecutive integers.
    static void checksum(int n)
    {
        // if n is multiple of 5
        if (n % 5 == 0)
            Console.WriteLine( (n / 5 - 2)
                + " " + (n / 5 - 1) + " "
                + (n / 5) + " " + (n / 5
                + 1 ) + " " + (n / 5 + 2));
     
        // else print "-1".
        else
            Console.WriteLine( "-1");
    }
     
    // Driver Program
    public static void Main ()
    {
        int n = 15;
        checksum(n);
    }
}
 
// This code is contributed by anuj_67.

Java脚本

// javascript Program to check if a number can
// be expressed as sum of five consecutive
// integer.
 
    // function to check if a number can
    // be expressed as sum of five
    // consecutive integers.
    function checksum(n) {
        // if n is multiple of 5
        if (n % 5 == 0)
            document.write((n / 5 - 2) + " "
            + (n / 5 - 1) + " " + (n / 5)
            + " " + (n / 5 + 1) + " " + (n / 5 + 2));
 
        // else prvar "-1".
        else
            document.write("-1");
    }
 
    // Driver Program
     
        var n = 15;
        checksum(n);
 
// This code is contributed by todaysgaurav

输出：

1 2 3 4 5