< C++ Programming < Exercises
#include <iostream>
#include <cmath>
using namespace std;
int sum (int,int);
int main ()
{
int x, y;
cout<<"enter first number: ";
cin>> x;
cout<<"enter second number: ";
cin>>y;
cout<<"Sum of these two :"<<sum(x,y);{
return 0;
}
int sum(int a, int b) {
int c = a+b;
return c;
}
EXERCISE 1
Write a program that performs arithmetic division. The program will use two integers, a and b (obtained by the user) and will perform the division a/b, store the result in another integer c and show the result of the division using cout. In a similar way, extend the program to add, subtract, multiply, do modulo and power using integers a and b. Modify your program so that when it starts, it asks the user which type of calculation it should do, then asks for the 2 integers, then runs the user selected calculation and outputs the result in a user friendly formatted manner.
Sample run:
Enter two numbers: 3 12 The sum is 15.
Solution
Solution #1
#include <iostream>
using namespace std;
int AddTwo (int addend1, int addend2) {
return addend1 + addend2;
}
int main () {
int number1, number2, sum;
cout << "Enter two integers:\n";
cin >> number1 >> number2;
sum = AddTwo(number1, number2);
cout << "\nThe sum is " << sum << ".";
}
Solution #2
//by blazzer12
//Input two values. Call a function that returns the sum of the values.
#include<iostream>
using namespace std;
int getSum(int, int);
int main()
{
int number1, number2, sum;
//get values
cout<<"Enter two integers to add"<<endl;
cout<<"Number1 :";
cin>>number1;
cout<<"Number2 :";
cin>>number2;
//call getSum() and store result in sum
sum = getSum(number1, number2);
//print result
cout<<number1<<" + "<<number2<<" = "<<sum;
}
int getSum(int addend1, int addend2)
{
return addend1 + addend2;
}
The solution in C.
#include <stdio.h>
int add(int, int);
int main()
{
int a, b, sum;
printf("A: ");
scanf("%d", &a);
printf("B: ");
scanf("%d", &b);
sum = add(a, b);
printf("The sum of %d and %d is %d.\n", a, b, sum);
return 0;
}
int add(int a, int b)
{
return a + b;
}
// Another solution:
# include <iostream>
using namespace std;
int sum (int number1, int number2);
int number1;
int number2;
int main()
{
cout<<"Give me a number amigo: ";
cin>>number1;
cout<<"Give me another number dude: ";
cin>>number2;
cout<<"The sum of "<<number1<<" and "<<number2<<" is: "<<sum(number1,number2)<<"."<<endl;
return 0;
}
int sum (int number1, int number2)
{
return number1+number2;
}
// by neuroalchemist
EXERCISE 2
Basically the same as exercise 1, but this time, the function that adds the numbers should be void, and takes a third, pass by reference parameter; then puts the sum in that.
Solution
Solution #1
#include <iostream>
using namespace std;
void AddTwo (int addend1, int addend2, int &sum) {
sum = addend1 + addend2;
}
int main () {
int number1, number2, sum;
cout << "Enter two integers:\n";
cin >> number1 >> number2;
AddTwo(number1, number2, sum);
cout << "\nThe sum is " << sum << ".";
return 0;
}
Solution #2
//by blazzer12
//adds two integers using a "pass by reference" type function call.
#include <iostream>
using namespace std;
void addNum(int, int, int&);
int main()
{
int number1,number2,sum;
//get values;
cout<<"Enter two integers to add"<<endl;
cout<<"Enter Number 1: ";
cin>>number1;
cout<<"Enter Number 2: ";
cin>>number2;
//call addNum to add the numbers
addNum(number1, number2, sum);
//print sum
cout<<number1<<" + "<<number2<<" = "<<sum;
return 0;
}
void addNum(int addend1, int addend2, int &sum)
{
sum = addend1 + addend2;
}
Solution #3
#include<iostream>
using namespace std;
void add(int,int);
int sum=0;
int main(){
cout<<"Please input the first number you wanna add:"<<endl;
int num1;
cin>>num1;
cout<<"Please input the second number you wanna ad"<<endl;
int num2;
cin>>num2;
add(num1,num2);
cout<<"The sum of these two number:"<<num1<<"&"<<num2<<" is:"<<sum<<endl;
return 0;
}
void add(int a,int b){
sum=a+b;
}
EXERCISE 3
Write a recursive function that finds the #n integer of the Fibonacci sequence. Then build a minimal program to test it. For reference see Wikipedia:Fibonacci number.
For any possible natural number "n", the following applies fib(n+2) = fib(n+1) + fib(n) Also, the following are predefined fib(0) = 0 fib(1) = 1
Solution
Solution #1
#include <iostream>
using namespace std;
unsigned fib(unsigned n);
int main()
{
// Printing the first 20 Fibonacci sequence values
for (unsigned i = 0; i < 20; i++){
cout << "fib(" << i << ") = " << fib(i) << endl;
}
}
unsigned fib(unsigned n)
{
if (n < 2)
return n;
return fib(n-2) + fib(n-1);
}
EXERCISE 4
Basically the same as exercise 3, although this time you mustn't use recursion.
Solution
Solution #1
#include <iostream>
using namespace std;
unsigned fib(unsigned n);
int main()
{
// Printing the first 20 Fibonacci sequence values
for (unsigned i = 0; i < 20; i++){
cout << "fib(" << i << ") = " << fib(i) << endl;
}
}
unsigned fib(unsigned n)
{
if (n < 2)
return n;
unsigned prev1 = 0;
unsigned prev2 = 1;
for (unsigned i = 0; i <= n-2; i++){
unsigned temp = prev1 + prev2;
// Just doing a rotation of values, since only the last two are needed
prev1 = prev2;
prev2 = temp;
}
return prev2;
}
For extra exercise, give a big number (like 1000000) to both exercise 3 and 4 solutions and compare the execution times. Ponder on the results ;)
EXERCISE 5
Create a calculator that takes a number, a basic math operator (+,-,*,/,^), and a second number all from user input, and have it print the result of the mathematical operation. The mathematical operations should be wrapped inside of functions.
Solution
Solution #1
Hammad city university
#include<iostream>
using namespace std;
void calculator(int num, int num2, int result);
void calculator(int num,int num2,int result)
{
char op;
cout<<"\n Calculator:- \nEnter Number: " ;
cin>>num;
cout<<"Enter operator +,-,*,/,^ : ";
cin >>op;
cout<<"Enter second number: " ;
cin>>num2;
if(op=='+')result=num+num2;
if(op=='-')result=num-num2;
if(op=='*')result=num*num2;
if(op=='/')result=num/num2;
if(op=='^')result=num^num2;
cout<<"result: "<<result<<"\n";
}
main()
{
int a,b,c;
calculator(a,b,c);
return 0;
}
}
Solution #2
#include <iostream>
float add(float a, float b)
{
return a + b;
}
float sub(float a, float b)
{
return a - b;
}
float mul(float a, float b)
{
return a * b;
}
float div(float a, float b)
{
if(b != 0)
{
return a / b;
}
std::cout << "Error: division by zero.\n";
return 0;
}
float pow(float a, float b)
{
return pow(a, b);
}
float mod(float a, float b)
{
return fmod(a, b);
}
void calc()
{
float a, b;
char op;
std::cout << "Enter a #: ";
std::cin >> a;
std::cout << "Enter an operator (+, -, *, /, ^, %): ";
std::cin >> op;
std::cout << "Enter a second #: ";
std::cin >> b;
switch(op)
{
case '+':
std::cout << "Result: " << add(a, b);
break;
case '-':
std::cout << "Result: " << sub(a, b);
break;
case '*':
std::cout << "Result: " << mul(a, b);
break;
case '/':
std::cout << "Result: " << div(a, b);
break;
case '^':
std::cout << "Result: " << pow(a, b);
break;
case '%':
std::cout << "Result: " << mod(a, b);
break;
default:
std::cout << "Error: operator not valid.\n";
}
}
int main()
{
calc();
return 0;
}
Soln #3
// This program will do simple calculations involving + - * / and ^ //
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int x,y,sum;
cout << "Enter first number ";
cin >> x;
cout << endl;
cout << "Enter second number ";
cin >> y;
cout << endl;
cout << 1 << " +" << endl;
cout << 2 << " -" << endl;
cout << 3 << " *" << endl;
cout << 4 << " /" << endl;
cout << 5 << " ^" << endl << endl;
cout << "What math would you like to do? ";
cin >> sum;
cout << endl;
switch (sum){
case 1:
sum = x + y;
cout << "The answer to your addition is " << sum << endl;
break;
case 2:
sum = x - y;
cout << "The answer to your subtraction is " << sum << endl;
break;
case 3:
sum = x * y;
cout << "The answer to your multiplication is " << sum<< endl;
break;
case 4:
sum = x / y;
cout << "The answer to your division is " << sum << endl;
break;
case 5:
sum = pow(x,y);
cout << "The answer to your power function is " << sum << endl;
break;
default:
cout << "You have entered an invalid option " << endl;
break;
}
return 0;
}
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