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

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

Sum of Consecutive Odd Numbers II

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

Input

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

Output

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

Input SampleOutput Sample

7

4 5

13 10

6 4

3 3

3 5

3 4

3 8

0

11

5

0

0

0

12

                                                          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 / BEE 1099 Solution in C :   

  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;
}

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