Beecrowd 1045 Triangle Types solution

Beecrowd 1045 Triangle Types solution in C, C++, Python

Beecrowd 1045 Triangle Types solution Question BEE 1045 – Triangle Types Solution in C #include <stdio.h> int main() { double a,b,c,temp,i,j; scanf(“%lf%lf%lf”,&a,&b,&c); 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){ printf(“NAO FORMA TRIANGULO\n”); } else{ if(j==i){ printf(“TRIANGULO RETANGULO\n”); } if(j>i){ printf(“TRIANGULO OBTUSANGULO\n”); } if(j<i){ printf(“TRIANGULO ACUTANGULO\n”); } if(a==b&&b==c){ printf(“TRIANGULO EQUILATERO\n”); } else if(a==b||b==c||c==a){ printf(“TRIANGULO …

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Beecrowd 1046 solution

Beecrowd 1046 solution || URI РBEECROWD РBEE Online Judge Solution  1046-Game Time || Uri 1046 solution in C, C++

submit link 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 …

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Beecrowd 1020 Solution

Beecrowd 1020 Solution in C, C++ & Python

URI – BEECROWD – BEE Online Judge || Age In Days – URI – BEE 1020 Solution in C,C++, Python Question Beecrowd 1020 Solution in C #include <stdio.h> int main() { int d,y,m; scanf(“%d”,&d); y=0; m=0; y=d/365; d=d%365; m=d/30; d=d%30; printf(“%d ano(s)\n%d mes(es)\n%d dia(s)\n”,y,m,d); return 0; } Bee1020 Solution in C++ #include <iostream> using namespace …

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CSE Subject

CSE Subject || CSE subject list in Bangladesh || BUBT 2023 || Engineering in Computer Science

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 …

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Beecrowd 1019 solution

Beecrowd 1019 solution || URI – BEECROWD – BEE 1019 Time Conversion Solutions in C, C++, and Python – Comprehensive Guide

URI – BEECROWD – BEE 1019 Time Conversion Solutions in C, C++, and Python || Beecrowd 1019 solution Question Understand the challenge: Beecrowd 1019 solution 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 …

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Uri 1080 Solution

Uri 1080 Solution || BEE 1080 || Beecrowd1080 Highest and Position

Beecrowd1080 Highest and Position question Uri 1080 Solution || BEE 1080 ||Free Code Center Highest and Position Adapted by Neilor Tonin, URI  Brazil Timelimit: 1 Read 100 integer numbers. Print the highest read value and the input position. Input The input file contains 100 distinct positive integer numbers. Output Print the highest number read and the …

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C++ Program to Read an Amount and Find Number of Notes

C++ Program to Read an Amount and Find Number of Notes

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: As we all …

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Average In C

Program To Calculate Average In C

Introduction: 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 …

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If Else in C

Nested if else in C Programming Examples1

Nested if else in c programming examples #include <stdio.h> int main() { int s; printf (“enter your score =”); scanf(“%d”,&s); if(s>=80 && s<=100){ printf(“A+”); } else if(s>=75 && s<=79){ printf(“A= 3.75”); } else if(s>=70 && s<=74){ printf(“A-=3.50”); } else if(s>=65&&s<=69){ printf(“B+=3.25”); } else if(s>=60 && s<=64){ printf(“B=3.00”); } else if(s>=55 && s<=59){ printf(“B-=2.75”); } else …

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