# URI / BEE CROWD 1099 - Sum of Consecutive Odd Numbers II Solution in C,C++,Python | URI - BEECROWD - BEE 1099 Solution in C,C++,Python

beecrowd | 1099

# Adapted by Neilor Tonin, URI  Brazil

Timelimit: 1

URI / BEE CROWD 1099 - Sum of Consecutive Odd Numbers II Solution in C,C++,Python | URI - BEECROWD - BEE 1099 Solution in C,C++,Python:

Read an integer that is the number of test cases. Each test case is a line containing two integer numbers and Y. Print the sum of all odd values between them, not including and Y.

## Input

The first line of input is an integer that is the number of test cases that follow. Each test case is a line containing two integer and Y.

## Output

Print the sum of all odd numbers between X and Y.

 Input Sample Output Sample 74 513 106 43 33 53 43 8 011500012

Demonstration:

This program takes user inputs . Then check the conditions for each test case contains two integer number x and y and compare those two numbers if x is greater than y , then swapping in ascending order and calculating the sum of all odds numbers by using condition in loop.

if(number%2 != 0) then it's an odd number. Then we add all those odd numbers.
Finally print the result.

N.B: Don't copy paste the code as same. Just try to understand it and try yourself.

## URI / BEECROWD Online Judge 1099 Solve in C :

```#include<stdio.h>

int main(){

int n;
int x, y, aux;
int soma;

scanf("%d",&n);

while(1){
if(n == 0) break;
scanf("%d%d",&x,&y);

if(x > y){
aux = x;
x = y;
y = aux;
}
soma = 0;
for(int i = x+1; i < y; i++){
if(i%2 != 0) soma += i;
}
printf("%d\n",soma);
n--;
}
return 0;
}```