URI Online Judge Solution 1101 || Sequence of Numbers and Sum

URI Online Judge | 1101

Sequence of Numbers and Sum

Adapted by Neilor Tonin, URI  Brazil

Timelimit: 1

Read an undetermined number of pairs values M and N (stop when any of these values is less or equal to zero). For each pair, print the sequence from the smallest to the biggest (including both) and the sum of consecutive integers between them (including both).

Input

The input file contains pairs of integer values M and N. The last line of the file contains a number zero or negative, or both.

Output

For each pair of numbers, print the sequence from the smallest to the biggest and the sum of these values, as shown below.

Input Sample	Output Sample
5 2 6 3 5 0	2 3 4 5 Sum=14 3 4 5 6 Sum=18

URI Online Judge Solution 1101 || Sequence of Numbers and Sum in JAVA

import java.io.IOException;

import java.util.Scanner;

  

public class Main {

  

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

  

       int X, Y;

        Scanner input =new Scanner(System.in);

         

        while (((X = input.nextInt()) > 0 )&&((Y = input.nextInt()) > 0 )) {

            int sum = 0;

            if (X > Y) {

                for (Y = Y; Y <= X; Y++) {

                    sum+=Y;

                    System.out.print(Y+" ");

                }

                System.out.print("Sum="+sum+"\n");

            }else {

                for (X = X; X <= Y; X++) {

                    sum+=X;

                    System.out.print(X+" ");

                }

                System.out.print("Sum="+sum+"\n");

            }

        }

  

    }

  

}

URI Online Judge Solution 1101 || Sequence of Numbers and Sum in C

#include<stdio.h>
int main(){
    int X, Y;

    while (((scanf("%d", &X)) > 0 )&&(((scanf("%d", &Y)) > 0 ))) {
        int sum = 0;
        if (X > Y) {
            for (Y = Y; Y <= X; Y++) {
                sum+=Y;
                printf("%d ", Y);
            }
            printf("Sum=%d\n", sum);
        }else {
            for (X = X; X <= Y; X++) {
                sum += X;
                printf("%d ", X);
            }
            printf("Sum=%d\n", sum);
        }
    }
}

URI Main Question link
Download dropbox code

Some Description about the problem:

Using a simple for loop I've just solve the problem.
Look in the line

for (Y = Y; Y <= X; Y++) {

                    sum+=Y;

                    System.out.print(Y+" ");

                }

Here from two input range I've calculated the summation.

Note : 

It is hardly required that you have to give all of the format what user can give the input. That means look-

 if else statement