## Beecrowd 1095 solution

## Sequence IJ 1

### Introduction

In this article, we will be discussing the URI Bee Crowd 1095 Sequence IJ 1 solution in C/C++/Python. We will provide a detailed explanation of the problem, walk you through the solution step by step, and provide code snippets in C/C++/Python to help you implement the solution. By the end of this article, you will have a better understanding of the problem and be able to solve it with ease.

### Problem Description

The URI Bee Crowd 1095 Sequence IJ 1 problem requires us to find the sum of all even numbers between two integers I and J, inclusive. The values of I and J will be given as input, and we need to output the sum of all even numbers between I and J. If I or J is an odd number, we need to consider the next even number as the starting or ending point.

## Implementation in C

#include<stdio.h> int main() { int i, j, sum = 0; scanf("%d%d", &i, &j); if(i%2 == 1) i++; for(; i <= j; i+=2) { sum += i; } printf("%d\n", sum); return 0; }

## Implementation in Cpp

#include <iostream> using namespace std; int main() { // cout<<"I=1"<<" "<<"j=60"<<endl; for(int i=1,j=60;i<15,j>=0;i+=3,j-=5){ cout<<"I="<<i <<" "<<"J="<<j<<endl; } return 0; }

## Beecrowd 1095 solution in Python

## Implementation in Python

i, j = map(int, input().split()) if i%2 == 1: i += 1 print(sum(range(i, j+1, 2)))

## Conclusion

In conclusion, we have provided a detailed explanation of the URI Bee Crowd 1095 Sequence IJ 1 problem and provided solutions in C/C++/Python. By implementing the solutions provided in this article, you will be able to solve the problem with ease. We hope this article has been helpful, and if you have any further questions or comments, please feel free to leave them below.

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