How can I accept and average positive numbers? When a negative number is entered, it should terminate the loop and display the average (excluding the negative number).
#include <iostream>
using namespace std ;
int main () {
int x,counter=0,sum;
float avg;
while (x>0) {
if(x<0) {
sum+=x;
counter++;
continue;
} else if (x>0) {
cin>>x;
sum+=x;
counter ++;
}
}
avg=(float)sum/counter;
cout<<avg<<endl;
return 0 ;
}
#include <iostream>
using namespace std;
int main() {
int sum = 0, count = 0, input; // initialize all variables
while (true) { // loop forever (until break is reached below)
cin >> input; // get user input
if (input < 0) break; // if it's negative, exit loop
sum += input; // otherwise add it to the running sum
count++; // increase the count by 1
}
if (count > 0) cout << (double)sum / count; // if we have at least one number, display avg
else cout << "No numbers to average" << endl; // else complain.
return 0;
}
Note that this will fail if the user provides bad input. If you need it to handle bad input, see here about cin types.
The implementation that you have immediately adds the input to your sum.
int main () {
int x,counter=0,sum;
float avg;
while (x>0) {
cin >> x;
if (x>0) {
sum+=x;
counter ++;
}
}
avg=(float)sum/counter;
cout<<avg<<endl;
return 0 ;
}
This would allow you to check the input before adding to your total.
It is also important to mention to avoid dividing by zero if the user's first input is a negative number.
here is an improved version, that checks if loop exist without any positive integers entered to avoid divide by zero error
#include <iostream>
using namespace std ;
int main () {
int x;
float counter=0,sum;
float avg;
bool isPositive = true;
while ( isPositive) {
cin>>x;
if(x>0)
{
sum+=x;
counter ++;
}
else {
isPositive = false;
}
}
// if we divide by zero, an error will occur
if(counter > 0)
{
avg=(float)(sum/counter);
cout<<avg<<endl;
}
else cout << "Please enter positive numbers";
return 0 ;
}
Related
I was doing this program in which I am supossed to print gapful numbers all the way up to a specific value. The operations are correct, however, for some reason after printing a couple of values the program crashes, what can I do to fix this problem?
Here's my code:
#include<math.h>
#include<stdlib.h>
using namespace std;
void gapful(int);
bool gapCheck(int);
int main(){
int n;
cout<<"Enter a top number: ";
cin>>n;
gapful(n);
system("pause");
return 0;
}
void gapful(int og){
for(int i=0; i<=og; i++){
fflush(stdin);
if(gapCheck(i)){
cout<<i<<" ";
}
}
}
bool gapCheck(int n){
int digits=0;
int n_save,n1,n2,n3;
if(n<100){
return false;
}
else{
n_save=n;
while(n>10){
n/=10;
digits++;
}
digits++;
n=n_save;
n1=n/pow(10, digits);
n2=n%10;
n3=n1*10 + n2;
if(n%n3 == 0){
return true;
}
else{
return false;
}
}
}
I'm open to any suggestions and comments, thank you. :)
For n == 110, you compute digits == 3. Then n1 == 110 / 1000 == 0, n2 == 110 % 10 == 0, n3 == 0*10 + 0 == 0, and finally n%n3 exhibits undefined behavior by way of division by zero.
You would benefit from more functions. Breaking things down into minimal blocks of code which represent a single purpose makes debugging code much easier. You need to ask yourself, what is a gapful number. It is a number that is evenly divisible by its first and last digit. So, what do we need to solve this?
We need to know how many digits a number has.
We need to know the first digit and the last digit of the number.
So start out by creating a function to resolve those problems. Then, you would have an easier time figuring out the final solution.
#include<math.h>
#include <iostream>
using namespace std;
void gapful(int);
bool gapCheck(int);
int getDigits(int);
int digitAt(int,int);
int main(){
int n;
cout<<"Enter a top number: " << endl;
cin>>n;
gapful(n);
return 0;
}
void gapful(int og){
for(int i=1; i<=og; ++i){
if(gapCheck(i)){
cout<<i << '-' <<endl;
}
}
}
int getDigits(int number) {
int digitCount = 0;
while (number >= 10) {
++digitCount;
number /= 10;
}
return ++digitCount;
}
int digitAt(int number,int digit) {
int numOfDigits = getDigits(number);
int curDigit = 0;
if (digit >=1 && digit <= numOfDigits) { //Verify digit is in range
while (numOfDigits != digit) { //Count back to the digit requested
number /=10;
numOfDigits -=1;
}
curDigit = number%10; //Get the current digit to be returned.
} else {
throw "Digit requested is out of range!";
}
return curDigit;
}
bool gapCheck(int n){
int digitsN = getDigits(n);
if (digitsN < 3) { //Return false if less than 3 digits. Single digits do not apply and doubles result in themselves.
return false;
}
int first = digitAt(n,1) * 10; //Get the first number in the 10s place
int second = digitAt(n,digitsN); //Get the second number
int total = first + second; //Add them
return n % total == 0; //Return whether it evenly divides
}
so i want this code to be able to count the exact number of powers of 2 but it won't stop until i put odd number in console , any fixes to my code? Thanks a lot in advance
enter code here :
#include <iostream>
using namespace std;
int main()
{
int a;
int counter=0;
cin >> a;
while(true){
cin >> a;
if(a%2==1)
break;
a/=2;
counter=counter+1;
}
cout << counter;
return 0;
}
You were missing some things:
you were not taking inputs as you want.
your counting procedure was wrong.
Soln:
If a number n is a 2's power, then the and operation of n and
n - 1 must be 0. And in all other cases, the result is not 0. Say,
n = 4 (in binary it is 100)
n - 1 = 3 (in binary it is 11)
n & (n - 1) = 0
100 (4)
& 011 (3)
-----------
000 (0)
use this technique
#include <iostream>
using namespace std;
int main()
{
int tests, a;
int counter = 0;
cin >> tests;
for (int i = 0; i < tests; i++)
{
cin >> a;
if ((a & (a - 1)) == 0)
counter = counter + 1;
}
cout << counter;
return 0;
}
If you want to evaluate all inputs power of two, I would use a double as input in order to get those exponent less than zero (0.5, 0.25, etc.).
With this aim, due to the fact that doubles are expressed in double-precision floating-point format (as defined in IEEE 754-2008 standard), you only should check that the normalized fraction got after descomposing the number with std::frexp (https://en.cppreference.com/w/cpp/numeric/math/frexp) is equal to 0.5:
#include <cmath>
bool isPowerOfTwo(double a)
{
int exp;
return std::frexp(a, &exp) == 0.5;
}
Then the code:
#include <cmath>
#include <iostream>
bool isPowerOfTwo(double a)
{
int exp;
return std::frexp(a, &exp) == 0.5;
}
int main() {
unsigned counter = 0;
while (true) {
double input;
std::cin >> input;
if (!isPowerOfTwo(input)) {
break;
}
counter++;
}
std::cout << "Number of inputs power of 2: " << counter << std::endl;
return 0;
}
I am working on some recursion practice and I need to write a program that reverse the input of an integer
Example of input : cin >> 12345; The output should be 54321
but if that integer is negative the negative sign needs to be appended to only the first number.
Example of input : cin >> -1234; output -4321
I am having a hard time getting my program to adapt to the negative numbers. The way I have it set up if I run
Example of test : 12345 I get the right output 54321
So my recursion and base are successful. But if I run a negative I get
Example of test : -12345 I get this for a reason I don't understand -5-4-3-2 1
#include<iostream>
using namespace std;
void reverse(int);
int main()
{
int num;
cout << "Input a number : ";
cin >> num;
reverse(num);
return 0;
}
void reverse(int in)
{
bool negative = false;
if (in < 0)
{
in = 0 - in;
negative = true;
}
if (in / 10 == 0)
cout << in % 10;
else{
if (negative == true)
in = 0 - in;
cout << in % 10;
reverse(in / 10);
}
}
To reverse a negative number, you output a - and then reverse the corresponding positive number. I'd suggest using recursion rather than state, like this:
void reverse(int in)
{
if (in < 0)
{
cout << '-';
reverse(-in);
}
else
{
// code to recursively reverse non-negative numbers here
}
}
Split the reverse function into two parts: the first part just prints - (if the input is negative) and then calls the second part, which is the recursive code you have. (You don't need any of the if (negative) ... handling any more, since the first part already handled it.)
Incidentally, if (bool_variable == true) ... is overly verbose. It's easier to read code if you say something like if (value_is_negative) ....
Your recursive function doesn't hold state. When you recurse the first time, it prints the '-' symbol but every time you send back a negative number to the recursion, it runs as if it is the first time and prints '-' again.
It's better to print '-' first time you see a negative number and send the rest of the number as a positive value to the recursion.
#include<iostream>
using namespace std;
void reverse(int);
int main()
{
int num;
cout << "Input a number : ";
cin >> num;
reverse(num);
return 0;
}
void reverse(int in)
{
bool negative = false;
if (in < 0)
{
in = 0 - in;
negative = true;
}
if (in / 10 == 0)
cout << in % 10;
else{
if (negative == true) {
cout << '-';
negative = false;
}
cout << in % 10;
reverse(in / 10);
}
}
int reverse(long int x) {
long int reversedNumber = 0, remainder;
bool isNegative = false;
if (x <0){
isNegative = true;
x *= -1;
}
while(x > 0) {
remainder = x%10;
reversedNumber = reversedNumber*10 + remainder;
x= x/10;
}
if (isNegative) {
if (reversedNumber > INT_MAX){
return 0;
}
else
return reversedNumber*(-1);
}
else
{
if (reversedNumber > INT_MAX){
return 0;
}
else
return reversedNumber;
}
}
I am writing a program to find the factorial of a user inputted number. My program works from, except for finding the factorial of 0. The requirement is that the factorial of 0 should output one, but I cannot think of a way to write this capability into the code without creating a special case for when 0 is entered. This is what I have so far
#include <iostream>
#include <cmath>
using namespace std;
int main() {
int startingNumber = 0;
double factorialize = NULL;
while(startingNumber != -1) {
cout << "Enter the numbr to factorial: ";
cin >> startingNumber;
factorialize = startingNumber;
for(int x=startingNumber-1;x>=1;x--) {
factorialize = factorialize*x;
}
cout << factorialize << endl;
factorialize = NULL;
}
return 0;
}
This outputs a factorial accurately for all cases except 0. Is there a way to do this that doesn't require a special case? I am thinking no because when I read about the reasons for why 0! is 1, it says that it is defined that way, in other words, you cannot reason your way into why it is 1. Just like x^0, 0! = 1 has a different logic as to why than why 2^2 is 4 or 2! = 2.
try this:
factorialize = 1;
for(int x=2; x<=startingNumber;x++)
factorialize *= x;
Try this:
for (unsigned int n; std::cin >> n; )
{
unsigned int result = 1;
for (unsigned int i = 1; i <= n; ++i) { result *= i; }
std::cout << n << "! = " << result << "\n";
}
You can change the result type a bit (unsigned long long int or double or long double), but ultimately you won't be able to compute a large number of factorials in hardware.
First of all I do not see how it can be calculated accurately, as you multiply startingNumber twice. So just fix the logic with:
factorialize = 1.0;
for(int x=startingNumber;x>=1;x--) {
factorialize = factorialize*x;
}
And it should calculate factorial properly as well as handling 0 the proper way.
Also you should not use NULL as initial value for double, it is for pointers.
There is a complete factorial of number program of C++ which includes the facility of factorial of positive number,negative and zero.
#include<iostream>
using namespace std;
int main()
{
int number,factorial=1;
cout<<"Enter Number to find its Factorial: ";
cin>>number;
if(number<0
)
{
cout<<"Not Defined.";
}
else if (number==0)
{
cout<<"The Facorial of 0 is 1.";
}
else
{
for(int i=1;i<=number;i++)
{
factorial=factorial*i;
}
cout<<"The Facorial of "<<number<<" is "<<factorial<<endl;
}
return 0;
}
You can read full program logic on http://www.cppbeginner.com/numbers/how-to-find-factorial-of-number-in-cpp/
The function listed below returns the factorial FASTER than any solution posted here to this date:
const unsigned int factorial(const unsigned int n)
{
unsigned int const f[13] = { 1,1,2,6,24,120,720,5040,40320,362880,3628800,39916800,479001600 };
return f[n];
}
I looks silly but it works for all factorials that fit into a 32-bit unsigned integer.
I am not sure whether I should ask here or programmers but I have been trying to work out why this program wont work and although I have found some bugs, it still returns "x is not a prime number", even when it is.
#include <iostream>
using namespace std;
bool primetest(int a) {
int i;
//Halve the user input to find where to stop dividing to (it will remove decimal point as it is an integer)
int b = a / 2;
//Loop through, for each division to test if it has a factor (it starts at 2, as 1 will always divide)
for (i = 2; i < b; i++) {
//If the user input has no remainder then it cannot be a prime and the loop can stop (break)
if (a % i == 0) {
return(0);
break;
}
//Other wise if the user input does have a remainder and is the last of the loop, return true (it is a prime)
else if ((a % i != 0) && (i == a -1)) {
return (1);
break;
}
}
}
int main(void) {
int user;
cout << "Enter a number to test if it is a prime or not: ";
cin >> user;
if (primetest(user)) {
cout << user << " is a prime number.";
}
else {
cout << user<< " is not a prime number.";
}
cout << "\n\nPress enter to exit...";
getchar();
getchar();
return 0;
}
Sorry if this is too localised (in which case could you suggest where I should ask such specific questions?)
I should add that I am VERY new to C++ (and programming in general)
This was simply intended to be a test of functions and controls.
i can never be equal to a - 1 - you're only going up to b - 1. b being a/2, that's never going to cause a match.
That means your loop ending condition that would return 1 is never true.
In the case of a prime number, you run off the end of the loop. That causes undefined behaviour, since you don't have a return statement there. Clang gave a warning, without any special flags:
example.cpp:22:1: warning: control may reach end of non-void function
[-Wreturn-type]
}
^
1 warning generated.
If your compiler didn't warn you, you need to turn on some more warning flags. For example, adding -Wall gives a warning when using GCC:
example.cpp: In function ‘bool primetest(int)’:
example.cpp:22: warning: control reaches end of non-void function
Overall, your prime-checking loop is much more complicated than it needs to be. Assuming you only care about values of a greater than or equal to 2:
bool primetest(int a)
{
int b = sqrt(a); // only need to test up to the square root of the input
for (int i = 2; i <= b; i++)
{
if (a % i == 0)
return false;
}
// if the loop completed, a is prime
return true;
}
If you want to handle all int values, you can just add an if (a < 2) return false; at the beginning.
Your logic is incorrect. You are using this expression (i == a -1)) which can never be true as Carl said.
For example:-
If a = 11
b = a/2 = 5 (Fractional part truncated)
So you are running loop till i<5. So i can never be equal to a-1 as max value of i in this case will be 4 and value of a-1 will be 10
You can do this by just checking till square root. But below is some modification to your code to make it work.
#include <iostream>
using namespace std;
bool primetest(int a) {
int i;
//Halve the user input to find where to stop dividing to (it will remove decimal point as it is an integer)
int b = a / 2;
//Loop through, for each division to test if it has a factor (it starts at 2, as 1 will always divide)
for (i = 2; i <= b; i++) {
//If the user input has no remainder then it cannot be a prime and the loop can stop (break)
if (a % i == 0) {
return(0);
}
}
//this return invokes only when it doesn't has factor
return 1;
}
int main(void) {
int user;
cout << "Enter a number to test if it is a prime or not: ";
cin >> user;
if (primetest(user)) {
cout << user << " is a prime number.";
}
else {
cout << user<< " is not a prime number.";
}
return 0;
}
check this out:
//Prime Numbers generation in C++
//Using for loops and conditional structures
#include <iostream>
using namespace std;
int main()
{
int a = 2; //start from 2
long long int b = 1000; //ends at 1000
for (int i = a; i <= b; i++)
{
for (int j = 2; j <= i; j++)
{
if (!(i%j)&&(i!=j)) //Condition for not prime
{
break;
}
if (j==i) //condition for Prime Numbers
{
cout << i << endl;
}
}
}
}
main()
{
int i,j,x,box;
for (i=10;i<=99;i++)
{
box=0;
x=i/2;
for (j=2;j<=x;j++)
if (i%j==0) box++;
if (box==0) cout<<i<<" is a prime number";
else cout<<i<<" is a composite number";
cout<<"\n";
getch();
}
}
Here is the complete solution for the Finding Prime numbers till any user entered number.
#include <iostream.h>
#include <conio.h>
using namespace std;
main()
{
int num, i, countFactors;
int a;
cout << "Enter number " << endl;
cin >> a;
for (num = 1; num <= a; num++)
{
countFactors = 0;
for (i = 2; i <= num; i++)
{
//if a factor exists from 2 up to the number, count Factors
if (num % i == 0)
{
countFactors++;
}
}
//a prime number has only itself as a factor
if (countFactors == 1)
{
cout << num << ", ";
}
}
getch();
}
One way is to use a Sieving algorithm, such as the sieve of Eratosthenes. This is a very fast method that works exceptionally well.
bool isPrime(int number){
if(number == 2 || number == 3 | number == 5 || number == 7) return true;
return ((number % 2) && (number % 3) && (number % 5) && (number % 7));
}