#include <bits/stdc++.h>
using namespace std;
int main() {
double a, b, c, temp, i, j;
cin >> a >> b >> c;
if (a < b) {
temp = a;
a = b;
b = temp;
}
if (a < c) {
temp = a;
a = c;//Beecrowd 1045 Triangle Types solution
c = temp;
}
if (b < c) {
temp = b;
b = c;
c = temp;
}
i = b * b + c * c;
j = a * a;
if (a >= b + c) {
cout << "NAO FORMA TRIANGULO" <<endl;
} else {
if (j == i) {
cout << "TRIANGULO RETANGULO" << endl;
}
if (j > i) {
cout << "TRIANGULO OBTUSANGULO" << endl;
}
if (j < i) {
cout << "TRIANGULO ACUTANGULO" <<endl;
}
if (a == b && b == c) {
cout << "TRIANGULO EQUILATERO" <<endl;
} else if (a == b || b == c || c == a) {
cout << "TRIANGULO ISOSCELES" <<endl;
}
}
return 0;
}
BEE 1045 – Triangle Types Solution in Python
def main():
a, b, c = map(float, input().split())
temp = 0
i = 0
j = 0
if a < b:
temp = a
a = b
b = temp
if a < c:
temp = a
a = c
c = temp
if b < c:
temp = b
b = c
c = temp
i = b * b + c * c
j = a * a
if a >= b + c:
print("NAO FORMA TRIANGULO")
else:
if j == i:
print("TRIANGULO RETANGULO")
if j > i:
print("TRIANGULO OBTUSANGULO")
if j < i:
print("TRIANGULO ACUTANGULO")
if a == b and b == c:
print("TRIANGULO EQUILATERO")
elif a == b or b == c or c == a:
print("TRIANGULO ISOSCELES")
if __name__ == "__main__":
main()
Beecrowd 1046 solution || URI – BEECROWD – BEE 1046 Solution in C, C++
Question
Game Time
Read the start time and end time of a game, in hours. Then calculate the duration of the game, knowing that the game can begin in a day and finish in another day, with a maximum duration of 24 hours. The message must be printed in portuguese “O JOGO DUROU X HORA(S)” that means “THE GAME LASTED X HOUR(S)”
Input
Two integer numbers representing the start and end time of a game.
Output
Print the duration of the game as in the sample output.Beecrowd 1046
Input Sample
Output Sample
16 2
O JOGO DUROU 10 HORA(S)
0 0
O JOGO DUROU 24 HORA(S)
2 16
O JOGO DUROU 14 HORA(S)
Beecrowd/Uri 1046 solution in C
#include <stdio.h>
int main() {
int a,b;
scanf("%d%d",&a,&b);
if(a==b){
printf("O JOGO DUROU 24 HORA(S)\n");
}
if(b>a){
printf("O JOGO DUROU %d HORA(S)\n",b-a);
}
if(a>b){
printf("O JOGO DUROU %d HORA(S)\n",(b+24)-a);//Beecrowd 1046 solution
}
return 0;
}
Beecrowd/Uri 1046 solution in C++
#include <iostream>
using namespace std;
int main() {
int a, b;
std::cin >> a >> b;
if (a == b) {
cout << "O JOGO DUROU 24 HORA(S)" <<endl;
} else if (b > a) {
cout << "O JOGO DUROU " << b - a << " HORA(S)" <<endl;
} else {
cout << "O JOGO DUROU " << (b + 24) - a << " HORA(S)" <<endl;
}
return 0;
}
#include <iostream>
using namespace std;
int main() {
int d, y, m;
cin >> d;
y = 0;
m = 0;
y = d / 365;
d = d % 365;
m = d / 30;
d = d % 30;
cout << y << " ano(s)\n" << m << " mes(es)\n" << d << " dia(s)\n";
return 0;
}
Bee1020 Solution in Python
d = int(input())
y = 0
m = 0
y = d // 365
d = d % 365
m = d // 30
d = d % 30
print(f"{y} ano(s)\n{m} mes(es)\n{d} dia(s)")
CSE Subject List in Bangladesh || CSE Full Course According to BUBT
BUBT is 163 credits consisting of 120 theory credits 37 lab credits and a capstone project of 6 credits.
FIRST-YEAR
CSE Subject list 1st Semester :
Course Title.
Credits
Course No
Structured Programming Language
3
CSE 101
Structured Programming Language Lab
1.5
CSE 102
Electrical Technology
3
CSE 107
Electrical Technology Lab
1.5
CSE 108
Differential and Integral Calculus
3
MAT111
Physics
3
PHY
English Language-I
3
ENG101
Principles of Economics
3
ECO
TotalCredit1st semester20 credit
CSE Subject 2nd Semester Full Course :
Course Title.
Credits
Object-Oriented Programming Language
3
Object-Oriented Programming Language Lab
1.5
Electronic Devices and Circuits
3
Electronic Devices and Circuits Lab
1.5
Statistics
3
Discrete Mathematics
3
Co-Ordinate Geometry and Vector Calculus
3
English Language-II
3
Software Development I
0.75
TotalCredit2nd semester21.75 credit
SECOND YEAR
CSE 3rd Semester Full Course :
Course Title.
Credits
Data Structures
3
Data Structures Lab
1.5
Digital Logic Design
3
Data Structures Lab
1.5
Accounting Fundamentals
2
Theory of Computing & Automata Theory
3
Linear Algebras and Differential Equations
3
Data Communication
3
TotalCredit3rd semester20 credit
CSE 4th Semester Full Course :
Course Title.
Credits
Course No
Algorithms
3
CSE240
AlgorithmsLAB
1.5
CSE241
Database Systems
3
CSE207
Database Systems LAB
1.5
CSE208
Complex Variable and Fourier Analysis
3
MAT231
Compiler Design
3
CSE323
Compiler Design LAB
.75
CSE324
Numerical Analysis Lab
.75
CSE224
Computer Architecture
3
CSE215
Software Development II
.75
CSE200
TotalCredit4th semester20.25 credit
3rd and 4th Year
COURSE NO
COURSE TITLE
CREDIT
PREREQUISITE
CSE 300
Software Development III
0.75
CSE 207
CSE 309
Operating Systems
3
CSE 111
CSE 310
Operating Systems Lab
1.5
CSE 111
CSE 313
Mathematical Analysis for Computer Science
3
STA 231
CSE 315
Microprocessor and Interfacing
3
CSE 215
CSE 316
Microprocessor and Interfacing Lab
1.5
CSE 215
CSE 317
System Analysis and Design
3
CSE 207
CSE 318
System Analysis and Design Lab
1.5
CSE 207
CSE 319
Computer Network
3
CSE 209
CSE 320
Computer Network Lab
1.5
CSE 209
CSE 323
Compiler Design
3
CSE 213
CSE 324
Compiler Design Lab
0.75
CSE 213
CSE 327
Software Engineering
3
CSE 317
CSE 328
Software Engineering Lab
0.75
CSE 317
CSE 331
Advanced Programming
3
CSE 121
CSE 332
Advanced Programming Lab
1.5
CSE 121
CSE 341
Computer Graphics
3
CSE 241
CSE 342
Computer Graphics Lab
0.75
CSE 241
CSE 351
Artificial Intelligence and Expert System
3
CSE 103
CSE 352
Artificial Intelligence and Expert System Lab
1.5
CSE 103
CSE 400
Software Development IV
0.75
CSE 300
CSE 407
Project Management and Professional Ethics
2
CSE 411
Digital Electronics and Pulse Technique
3
CSE 205
CSE 412
Digital Electronics and Pulse Technique Lab
1.5
CSE 205
CSE 413
Cyber Security and Digital Forensic
3
CSE 319
CSE 414
Cyber Security and Digital Forensic Lab
1.5
CSE 319
CSE 417
Distributed Database Management Systems
3
CSE 207
CSE 418
Distributed Database Management Systems Lab
1.5
CSE 207
CSE 425
Microcontroller and Embedded Systems
3
CSE 315
CSE 426
Microcontroller and Embedded Systems Lab
1.5
CSE 315
CSE 431
Communication Engineering
3
CSE 209
CSE 433
Fiber Optics Communication
3
CSE 209
CSE 435
Network Security
3
CSE 319
CSE 437
Digital Signal Processing
3
CSE 411
CSE 438
Digital Signal Processing Lab
1.5
CSE 411
CSE 439
Wireless Networking
3
CSE 319
CSE 440
Wireless Networking Lab
1.5
CSE 319
CSE 441
Switching and Routing
3
CSE 319
CSE 442
Switching and Routing Lab
1.5
CSE 319
CSE 443
Delay Tollerant Network
3
CSE 319
CSE 444
Delay Tollerant Network Lab
1.5
CSE 319
CSE 445
Introduction to Cryptography
3
CSE 319
CSE 446
Introduction to Cryptography Lab
1.5
CSE 319
CSE 447
Mobile Communication
3
CSE 209
CSE 448
Mobile Communication Lab
1.5
CSE 209
CSE 449
Network Administration
3
CSE 319
CSE 450
Network Administration Lab
1.5
CSE 319
CSE 451
Software Project Management
3
CSE 327
CSE 453
Software Testing & Quality Assurance
3
CSE 327
CSE 455
Software Security & Authentication
3
CSE 327
CSE 457
Web Database Programming
3
CSE 207
CSE 458
Web Database Programming Lab
1.5
CSE 207
CSE 459
Visual Programming
3
CSE 207
CSE 460
Visual Programming Lab
1.5
CSE 207
CSE 461
Software Architecture
3
CSE 327
CSE 462
Software Architecture Lab
1.5
CSE 327
CSE 463
Software Design Pattern
3
CSE 327
CSE 464
Software Design Pattern Lab
1.5
CSE 327
CSE 465
Machine Learning
3
CSE 351
CSE 467
Pattern Recognition
3
CSE 351
CSE 469
Basic Graph Theory
3
CSE 341
CSE 471
Computational Geometry
3
CSE 213
CSE 473
Simulation and Modeling
3
CSE 351
CSE 474
Simulation and Modeling Lab
1.5
CSE 351
CSE 475
Data Mining
3
CSE 351
CSE 476
Data Mining Lab
1.5
CSE 351
CSE 477
Neural Network and Fuzzy Systems
3
CSE 351
CSE 478
Neural Network and Fuzzy Systems Lab
1.5
CSE 351
CSE 479
VLSI Design
3
CSE 351
CSE 480
VLSI Design Lab
1.5
CSE 351
CSE 481
Decision Support System
3
CSE 351
CSE 482
Decision Support System Lab
1.5
CSE 351
CSE 483
Knowledge Engineering
3
CSE 351
CSE 484
Knowledge Engineering Lab
1.5
CSE 351
CSE 485
Parallel Processing
3
CSE 319
CSE 487
Robotics and Computer Vision
3
CSE 319
CSE 489
Fault Tolerant System
3
CSE 319
CSE 491
Interfacing & Peripherals
3
CSE 492
Interfacing & Peripherals Lab
1.5
CSE 493
Digital System Design
3
CSE 411
CSE 494
Degital System Design Lab
1.5
CSE 411
CSE 498
Project/Thesis
4
CSE 499
Capstone Project
4
CSE 300
Engineering in Computer Science and Engineering degree at BUBT is 163 credits consisting of 120 theory credits and 37 lab credits and a capstone project of 6 credits. The assignment of credits to a theoretical course follows a different rule from that of a sessional course.CSE Subject list.
Beecrowd 1019 solution presents the classic scheduling challenge: time shifting. In the challenge, times in seconds are converted to hours, minutes and seconds. This seemingly simple task requires a deep understanding of time distribution and variables. Our team of experts cracked the challenge and developed a comprehensive solution that not only solves the problem but also provides insight into the underlying concepts.
The essence of the time change
Time is a universal constant, but its representation varies from situation to situation. In programming, time is usually represented in seconds, and should be converted to a more human-readable format. This is where seasonal change becomes inevitable. Time conversion efficiency enables programmers to create applications that display time in a user-friendly manner, synchronize events, and perform robust calculations based on time intervals
Building the Solution: C, C++, Python Style
Our experts carefully developed a solution for URI Online Judge 1019 using three popular programming languages: C, C++ and Python. The solution not only provides a successful conversion but also demonstrates the versatility of these languages. Let’s take a closer look at each function:
URI – BEECROWD – BEE 1019 Time Conversion Solutions in C, C++, and Python ||Beecrowd 1019 solution
URI Online Judge 1019 Solve in C :
#include <stdio.h>
int main() {
int h,m,s;
scanf("%d",&s);
h=0;
m=0;
h=s/3600;
s=s%3600;
m=s/60;
s=s%60;
printf("%d:%d:%d\n",h,m,s);//Beecrowd 1019 solution
return 0;
}
INPUT: 556
OUTPUT: 0:9:16
URI Online Judge 1019 Solve in C++ :
#include<iostream>
using namespace std;
int main() {
int h,m,s;
cin>>s;
h=0;
m=0;
h=s/3600;
s=s%3600;
m=s/60;
s=s%60;
cout<<":"<<h<<":"<<m<<":"<<s<<endl;
return 0;
}
One crucial ability that cuts across language borders is time management. When this ability is honed, programmers can produce effective, user-friendly apps that faithfully depict the past. Our thorough guide offers complete solutions in C, C++, and Python for the URI Online Judge 1019 problem. For your benefit, we have also examined cutting-edge subjects and offered examples. You are equipped with this manual to tackle time-varying issues and hone your organizing abilities.
Start learning to program today with [Free Code Center]. Coding is fun!
C++ program to count all notes in a certain quantity – In this unique essay, we’ll go over the many C++programming techniques for counting all the notes in a certain quantity.
The following are the methods used in C++ programming to determine how many notes overall there are in a given quantity:
Using the Standard Procedure
Making Use of User-Defined Function
Employing Pointers
As we all know, with cash bundles, a number of notes combine to equal the exact quantity of money that the person in question needs.
Similar to that, this blog discusses the various methods for counting the number of notes in a certain amount. In a bundle, a single note’s multiples are multiplied.
The total number of notes in a particular quantity can be counted using a C++ program. Check out the whole code with example programs and sample outputs here. You can check out more introductory C++ programs here.
Thus, there are a number of ways to calculate notes in a given quantity in C++ programming, as follows:
C++ Program to Read an Amount and Find Number of Notes
Logic to count minimum number of denomination for given amount
There are numerous efficient algorithms available to address the given issue. To simplify things for this experiment, I utilized the greedy method.
Using descriptive logic, determine the least amount of denominations.
Amount entered by the user. Put it in a variable, like amt. If the amount is higher than 500, divide it by 500 to get the maximum number of notes needed, which is 500. Put the result of the division in a variable, such as note500 = amt / 500;. Divide the original sum by 500 notes, and then deduct the result. Execute note500 * 500 – amt.
For each note, repeat the previous step 200, 100, 50, 20, 10, 5, 2, and 1.
#include<iostream>
using namespace std;
class money{
public:
int note_size[9]={1000,500,200,100,50,20,10,5,2};
int note_count[9]={0};
money(int amount){
for (int i=0;i<9;i++){
if(amount>=note_size[i]){
note_count[i]=amount/note_size[i];
amount=amount%note_size[i];
}
}
cout<<"counted notes >>"<<endl;
for (int i=0;i<9;i++){
if(note_count[i]!=0){ cout<<note_size[i]<<" taka x"<<note_count[i]<<endl;
} } } };
int main(){ int amount;
cout<<"enter total amount : "; //C++ Program to Read an Amount and Find Number of Notes
cin >>amount;
money ob(amount);
return 0;}
C is one of the most potent and flexible programming languages available today. Any C program has to understand how to compute averages and other mathematical operations. The fundamental idea of enables programmers to efficiently examine and work with data.
We shall examine the typical world of C programming in this essay. We’ll talk about averages, how to calculate them, and how to successfully use them to planning. Whether you are a seasoned C programmer or just getting started, this article will provide you with insightful knowledge and a clear understanding of what average infeels are like. The sum of a group of numbers divided by the total number is the average. It can be characterized as
Average is equal to the total of all values divided by the total number of values.
Program To Calculate Average In C
Implementation
This algorithm’s implementation is shown below:
#include<stdio.h>
void main()
{
int i,n,sum=0 ;double aver=0;
printf("enter the number:");
scanf("%d",&n);
for(i=1;i<=10;i++){
sum=sum+i;
aver=sum/(double)10;
}
printf("The sum of 10 no is :%d\n",sum);
printf("The Average is :%lf",aver);
}
Input :Enter the number:6
Output: The sum of 6 no is :21 The Average is :2.100000
Implementation 2
Implementation of this algorithm is given below
#include <stdio.h>
int main() {
int data[] = {10, 20, 30, 40, 50}; // Sample data
int n = sizeof(data) / sizeof(data[0]);
int sum = 0;
for (int i = 0; i < n; i++) {
sum += data[i];
}
float average = (float)sum / n;
printf("The average is: %.2f", average);
return 0;
}