Hey people, this team we are going to do the second partial exam, so there’s the link for you, so you be able to check out the problems.
#1
#2
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#4
Enjoy
Fundamentals and Solutions. Together!
Hey people, this team we are going to do the second partial exam, so there’s the link for you, so you be able to check out the problems.
#1
#2
#3
#4
Enjoy
Hey people, we are in the finish line, so, let’s make a little more effort to finish this curse.
So, heré’s what we need to do:
Seems like a difficult one, but it isn’t, check this out:
If you have any question ask in the comments section below.
Let’s keep moving forward people
Last quiz everyone!!!!! jajajaja in this quiz I had to calculate the dot product of two vectors “seen in physics”, using arrays.
//CODE:
#include <iostream>
using namespace std;
double dot_product(double list1[], double list2[], int l1){
double dot=0;
for(int i=0; i<l1; i=i+1){
dot=dot+(list1[i]*list2[i]);
}
return dot;
}
int main(){
int l1;
double product;
cout<<“This program help to calculate the dot product of the lists of values given by the user.”<<endl;
cout<<“Please enter the values to the first list.”<<endl;
cout<<“How many caracters the lists have?”<<endl;
cin>>l1;
double list1[l1], list2[l1];
for(int i=0; i<l1; i=i+1){
cout<<“Type the ” <<i+1<<” “<<“value of the 1st list.”<<” “;
cin>>list1[i]; }
cout<<“Please enter the values to the second list.”<<endl;
for(int i=0; i<l1; i=i+1){
cout<<“Type the ” <<i+1<<” “<<“value of the 2nd list”<<” “;
cin>>list2[i]; }
product=dot_product(list1, list2, l1);
cout<<“The dot product of both lists is:”<<” “<<product<<endl;
return 0;
}
Write a function to calculate the greatest common denominator of two positive integers using Euclid’s algorithm.
//CODE:
#include <iostream>
using namespace std;
int gcd(int x, int y){
int r;
if (x==0){
return x;
}
while (y!=0){
r= y;
y= x%y;
x= r;
}
return r;
}
int main (){
int x,y;
cout<<“Please enter a number: ” <<endl;
cin>>x;
cout<<“Please enter a second number: ” <<endl;
cin>>y;
cout<<“The greatest common denominator between (” <<x <<” y ” <<y <<“) is = ” << gcd(x,y) <<endl;
return 0;
}
Determine if the word given by the user is a palindrome or not, by making the word backwards and comparing it to the original word with a simple “if condition”.
//CODE:
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
string word, fin;
string is_palindrome(string word)
{
string backwards;
cout << “Type a word: n”;
cin >> word;
int large = word.length();
for (int x = large – 1; x >= 0; x–)
{
backwards += word[x];
}
if (backwards == word)
{
fin = “Your word IS a palindrome”;
}
else
fin = “Your word IS NOT a palindrome”;
return fin;
}
int main()
{
cout << is_palindrome(fin);
}
Create a function called euler_calc with a single parameter precision. The value of precision is used to determine when to stop calculating. Your calculation will stop when the two consecutive values estimating e differ by less than precision
//CODE
#include <iostream>
#include <iomanip>
using namespace std;
double factorial(int n){
double fact = 1;
for (int i = 1; i <= n; i++){
if (n==0){
return 1;
}
else {
fact = fact * i;
}
}
return fact;
}
float euler_calc (float ac){
float e = 1.0;
for (int i=1; i<999; i++){
e = e + 1/(factorial(i));
}
cout<< fixed <<setprecision(ac) << e;
return e;
}
int main(){
float ac;
cout<<“How accurate you want the result? “;
cin>>ac;
cout<<“Here is your number: “<<endl;
euler_calc(ac);
cout<<endl;
return 0;
}
Here is the partial exam answered of course jajajaja so the exam just had four questions and I will list them in order.
QUESTION 1:
Write a function called triangles which receives a single parameter (int) which represents the size of a triangle as explained below. The function should print a triangle using loops (for or while). The only characters printed here are ‘T’ and the new-line character. The first line is length one, the middle line is length size and the last line is length one. The example below is for size 6.
T TT TTT TTTT TTTTT TTTTTT TTTTTT TTTTT TTTT TTT TT T
//CODE
// Ever Olivares
#include <stdio.h>
#include <iostream>
using namespace std;
int triangle (int x)
{
for (int i=1; i<=x ;i++)
{
for (int j=1; j<i+1; j++)
cout <<“T”; cout <<endl;
}
for (int i=x-1; i>=1; i–)
{
for (int j=1; j<i+1; j++)
cout <<“T”; cout <<endl;
}
return 0;
}
int main ()
{
int y;
std::cout<<“type the numbers of T you want” <<std::endl;
std::cin >> y;
triangle (y);
}
QUESTION 2:
Write a function called superpower that has two parameters of type long and returns a long which is first parameter raised to the power of the second, which is to say it returns ab So, superpower(3,4) would return 81.
//CODE:
// Ever Oliavres
#include <iostream>
#include <cmath>
void superpower(int num1, int num2){
int R;
R= pow(3,4);
std::cout << “The result is… “<<R << std::endl;
}
int main(int argc, char const *argv[]) {
int R;
superpower(3,4);
return 0;
}
QUESTION 3:
Write a function called fibonacci which receives a long “n” and returns a long which is the value of the nth number in the fibonacci series which is: 0,1,1,2,3,5,8,13,21,34,55,89…………So, fibonacci(0) would return 0. fibonacci(5) would return 5, fibonacci(8) would return
This one was cool but I do not know if it really work because it compiles and runs but do not count any words :(… anyway here it is.
//CODE
#include <iostream>
#include <string>
#include <fstream>
using namespace std;
int find_word(string word, string tipo){
string line;
int cont=0;
ifstream myfile(tipo);
if (myfile.is_open()){
while ( getline (myfile,line) ){
if(‘line’ == ‘word’){
cont=cont+1;
}
// cout << line << endl;
}
myfile.close();
}
else {
cout<<“CANNOT OPEN THE FILE”<<endl;
}
cout<<“The word…”<<word<<“… is used “<<cont<<” times.”<<endl;
return cont;
}
int main(int argc, char const *argv[]) {
string word, tipo;
int cont;
std::cout << “Please type the word you want to search.” << std::endl;
std::cin >> word;
std::cout << “Please enter the name of the file to search the word” << std::endl;
std::cin >> tipo;
find_word(word, tipo);
return 0;
}
This one in particular was a challenge… but with help everything is possible. Also for this WSQ I had to “add” a new library so I could work with bigger numbers it is called “BigIntegerLibrary.hh”.
//CODE
#include <iostream>
#include <string>
#include “BigIntegerLibrary.hh”
using namespace std;
BigInteger lychrel (BigInteger x){
string y = bigIntegerToString(x);
lychrel(y.begin(), y.end());
BigInteger lychrel = stringToBigInteger(y);
return lychrel;
}
int main() {
int low,up,number,j,k,p=0;
int l=0, n=0;
BigInteger z,count=0, sum=0;
cout << “Give me the lower bound”;
cin>>low;
cout<< “Give me the upper bound”;
cin>>up;
for(count=low;count<=up;count++){
for(BigInteger i=count; i<=count;i++){
z= lychrel(i); sum=y+count;}
if(count==z) {
n=n+1;
} else {
int times= 0;
BigInteger zeit = count;
z=lychrel(zeit);
while(zeit != z && times < 30){
zeit = zeit + z;
z = lychrel(zeit);
times=times+1;
}
if(times < 30){
p = p + 1;
} else {
cout << “Lychrel found ” << count << endl;
l = l +1;
}
}
}
cout <<“The palindrom is: ” << n <<endl;
cout<<“Lychrel doesn’t become palindrome in: “<< p <<endl;
cout<<“The lychrel is: “<< l <<endl;
return 0;
}