Trouble on even number

Sums of consecutive positive integers

Filename: Consecutive.java

Input file: consecutive.in

Consider the following positive integers: 5, 25, 29, 41, and 100. Each of these numbers can be expressed as a sum of n consecutive positive integers, where n > 1, as follows:

5 = 2 + 3

25 = 3 + 4 + 5 + 6 + 7

29 = 14 + 15

41 = 20 + 21

100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

1; 000; 000 = 7; 749 + … + 7; 876

Note that the decomposition is not unique. For example, observe that 25 is also equal to 12 + 13. However, 8, 32, and 1,024 are examples of numbers that cannot be expressed as sums of consecutive positive integers.

Your task is to determine if a positive integer can be expressed as a sum of consecutive positive integers, and to display the decomposition if that is possible.

INPUT FORMAT

The input consists of one or more lines with each line consisting of a positive integer.

OUTPUT FORMAT

The output consists of one or more lines with each line consisting of the given positive integer followed by the _rst and last term of its decomposition into a sum of consecutive positive integers. If the decomposition is not possible, then the line should display \n is not possible.", where n is the given positive integer. (Note: Any among the correct decompositions is acceptable).

SAMPLE INPUT

5

25

29

32

100

1000000

SAMPLE OUTPUT

5 2 3

25 3 7

29 14 15

32 is not possible.

100 9 16

1000000 7749 7876

import java.io.*;

public class Consecutive{

public static void main(String args[]) throws IOException{

BufferedReader inFile = new BufferedReader(new FileReader("consecutive.in"));
String str;
String num[ ]=new String[50];
int sum1=0,sum2=0, sum3=0, sum4=0;
int index=1, Soln=0;


while((str=inFile.readLine()) != null){
num=str.split(" ");
int num1 = Integer.parseInt(num[0]);


****if(num1%2==0){
System.out.println( num1 + "		" + sum3 +"   " + sum4);}**** im having trouble with this part.

else if(num1%2 != 0){
sum1=(num1/2);
sum2=(num1/2)+1;
System.out.println( num1 + "		" + sum1 +"   " + sum2);
		}
	      }
}
}