I was looking for solution to question what is the 10001st prime number.
And i am done with the code :
int main() {
long long listNumber[10001];
long position = 1, divider = 0;
listNumber[0] = 2;
while(listNumber[10000] == 0) {
divider = 0;
listNumber[position] = listNumber[position-1] + 1;
while(listNumber[divider] <= sqrt(listNumber[position])) {
if(listNumber[position] % listNumber[divider] == 0) {
listNumber[position]++;
divider = 0;
} else divider++;
}
position++;
}
cout << listNumber[10000] << endl;
return 0;
}
but the output is always change, i don't know why. Can you help me to figure it out?
Thank You.
You never initialize the array. That means its contents will be indeterminate and even reading that contents (like you do in the loop condition) leads to undefined behavior.
You need to initialize the array:
long long listNumber[10001] = {}; // Initialize all elements to zero
Related
#include<iostream>
using namespace std;
int main(){
int i = 1;
int sum;
int N;
cout << "Enter a number N: ";
cin >> N;
while(i<=N)
{
if(i%2 == 0)
{
sum = sum + i;
}
else
{
i = i + 1;
}
}
cout << sum;
}
This is to print the sum of all even numbers till 1 to N.
As I try to run the code, I am being asked the value of N but nothing is being printed ahead.
For starters the variable sum is not initialized.
Secondly you need to increase the variable i also when it is an even number. So the loop should look at least like
while(i<=N)
{
if(i%2 == 0)
{
sum = sum + i;
}
i = i + 1;
}
In general it is always better to declare variables in minimum scopes where they are used.
So instead of the while loop it is better to use a for loop as for example
for ( int i = 1; i++ < N; ++i )
{
if ( i % 2 == 0 ) sum += i;
}
while(i<=N)
{
if(i%2 == 0)
{
sum = sum + i;
}
else
{
i = i + 1;
}
}
Let's step through this. Imagine we're on the loop where i = 2 and you've entered N = 5. In that case...
while(i <= N)
2 <= 5 is true, so we loop
if(i%2 == 0)
2 % 2 == 0 is true, so we enter this branch
sum = sum + i;
Update sum, then head back to the top of the loop
while(i <= N)
Neither i nor N have changed, so 2 <= 5 is still true. We still loop
if(i%2 == 0)
2 % 2 == 0 is still true, so we enter this branch again...
Do you see what's happening here? Since neither i nor N are updated, you'll continue entering the same branch and looping indefinitely. Can you think of a way to prevent this? What would need to change?
Also note that int sum; means that sum will have a garbage value (it's uninitialized). If you want it to start at 0, you'll need to change that to
int sum = 0;
You're looping infinitly when i is even because you don't increase it.
Better option would be this if you want to use that while loop :
while(i<=N)
{
if(i%2 == 0)
sum = sum + i;
i=i+1;
}
cout << sum;
If you don't need to do anything when the condition is false, just don't use an else.
No loops are necessary and sum can be evaluated at compile time if needed too
// use unsigned, the whole excercise is pointless for negative numbers
// use const parameter, is not intended to be changed
// constexpr is not needed, but allows for compile time evaluation (constexpr all the things)
// return type can be automatically deduced
constexpr auto sum_of_even_numbers_smaller_then(const unsigned int n)
{
unsigned int m = (n / 2);
return m * (m + 1);
}
int main()
{
// compile time checking of the function
static_assert(sum_of_even_numbers_smaller_then(0) == 0);
static_assert(sum_of_even_numbers_smaller_then(1) == 0);
static_assert(sum_of_even_numbers_smaller_then(2) == 2);
static_assert(sum_of_even_numbers_smaller_then(3) == 2);
static_assert(sum_of_even_numbers_smaller_then(7) == 12);
static_assert(sum_of_even_numbers_smaller_then(8) == 20);
return 0;
}
int main(){
int input; //stores the user entered number
int sum=0; //stroes the sum of all even numbers
repeat:
cout<<"Please enter any integer bigger than one: ";
cin>>input;
if(input<1) //this check the number to be bigger than one means must be positive integer.
goto repeat; // if the user enter the number less than one it is repeating the entry.
for(int i=input; i>0; i--){ // find all even number from your number till one and than totals it.
if(i%2==0){
sum=sum+i;
int j=0;
j=j+1;
cout<<"Number is: "<<i<<endl;
}
}
cout<<endl<<"The sum of all even numbers is: "<<sum<<endl;}
Copy this C++ code and run it, it will solve your problem.
There are 2 problems with your program.
Mistake 1
The variable sum has not been initialized. This means that it has(holds) an indeterminate value. And using this uninitialized variable like you did when you wrote sum = sum + i; is undefined behavior.
Undefined behavior means anything1 can happen including but not limited to the program giving your expected output. But never rely on the output of a program that has undefined behavior.
This is why it is advised that:
always initialize built in types in local/block scope.
Mistake 2
The second problem is that you're not updating the value of variable i.
Solution
You can solve these problems as shown below:
int main(){
int i = 1;
int sum = 0; //INITIALIZE variable sum to 0
int N;
cout << "Enter a number N: ";
cin >> N;
while(i<=N)
{
if(i%2 == 0)
{
sum = sum + i;
}
i = i + 1; //update(increase i)
}
cout << sum;
}
1For more reading(technical definition of) on undefined behavior you can refer to undefined behavior's documentation which mentions that: there are no restrictions on the behavior of the program.
Basically i want to write a function that takes values from 0 to 6 and gives back a random assortment such as 2,3,4,5,0,1,6. Here is the code that i came up with. However the problem is that the integer prev (meaning previous) does not store all the old values of r (random number) and thus some values end up being repeated. How might i fix this?
int s(int b)
{
// b is 7
int h = b-1;
int prev = -1;// to store the previous r value
srand(time(0));
for (int i = 0; i < b; i++)
{
int r = rand()%(h - 0 + 1) + 0;
if (r != prev)
{
cout << r << endl;
prev = r;
}
else if (r == prev)
{
s(b);
}
}
return 0;
}
From the comments, this sounds more like a homework problem than a practical problem because you said "No arrays allowed". But I suppose it is an interesting problem.
Here's some code, in Java with only loops, if statements, and with no arrays, as required.
It outputs a random permutation of the set 0, 1, ..., N, shuffled with the Fisher-Yates algorithm.
void printRandom(int N) {
long used = 0;
for (int i = 0; i < N; i++) {
int randomIndex = ThreadLocalRandom.current().nextInt(N - Long.bitCount(used));
for (int j = 0; j < N; j++) {
if ((used & (1L << j)) == 0) {
if (randomIndex-- == 0) {
System.out.print(j + " ");
used = used | (1L << j);
break;
}
}
}
}
}
It is unfortunately limited to the size of a long on your system :)
I think the best way to solve this problem is by using an aux funtion that stores in a variable all the numbers printed until the moment, check if the new number is in the used numbers variable, if not add it to the variable (you can use strings? I know that they are arrays of char's but maybe you can)
Something like this:
function aux(int b, char *variables_printed, int iterations_left)
if (b = 0) then print variables_printed
else
int n = generate_random_number() %b
while (n in variables_printed)
n= (n+random_number) % b
variables_printed += n
aux(b, variables_printed, iterations_left-1)
And your other function:
function s(b)
if b < 0 return 0
else
char *variables_to_print
aux(b, variables_to_print, b)
If you can not use strings, you can do it with long as konsolas said.
The question:
Input data will give the number of test-cases in the first line.
Then test-cases themselves will follow, one case per line.
Each test-case describes an array of positive integers with value of 0 marking end. (this zero should not be included into calculations!!!).
Answer should contain average values for each array, rounded to nearest integer (see task on rounding), separated by spaces.
Problem:
Works fine but at third indice sum is assigned value of arrayInput and it messes everything up. Why does this happen and how can I fix it?
//araytest
#include<cmath>
#include<iostream>
using namespace std;
int main()
{
//var
int i = 0;
int array[13] = {};
//take in # arrays
cin >> i;
for(int x = 0; x<i; x++ )
{
//reset variables (for every array)
float arraySize = 0,
sum = 0, avg = 0;
int indice = 0,
arrayInput = 0;
while(cin >> arrayInput){
if(arrayInput == 0)
{
if(indice == 0)
{
arraySize = 1; /*if only 0 put in first indice
to prevent divide by 0 */
break;
}
else
{
arraySize = indice; // 0 doesn't count
break;
}
}
sum += arrayInput;
array[indice] = arrayInput;
arrayInput = 0;
indice++;
}
avg = round(sum/arraySize);
cout << avg << " ";
}
return 0;
}
First, like other people said, the array you used in this code is totally useless. It did nothing but save arrayinput.
Second, you let arraysize sum avg to be type float. However, arrayinput is assigned to be integer!! That means you never get result like this 2.xxx. So the type you declare for variables is meaningless. They should have same type declaration. I don't understand why you code does not work well. Because if you enter integer number, you wont get anything wrong. But it will crash if you give number like 2.xxx or x.xxx.
The program runs but it also spews out some other stuff and I am not too sure why. The very first output is correct but from there I am not sure what happens. Here is my code:
#include <iostream>
using namespace std;
const int MAX = 10;
int sum(int arrayNum[], int n)
{
int total = 0;
if (n <= 0)
return 0;
else
for(int i = 0; i < MAX; i ++)
{
if(arrayNum[i] % 2 != 0)
total += arrayNum[i];
}
cout << "Sum of odd integers in the array: " << total << endl;
return arrayNum[0] + sum(arrayNum+1,n-1);
}
int main()
{
int x[MAX] = {13,14,8,7,45,89,22,18,6,10};
sum(x,MAX);
system("pause");
return 0;
}
The term recursion means (in the simplest variation) solving a problem by reducing it to a simpler version of the same problem until becomes trivial. In your example...
To compute the num of the odd values in an array of n elements we have these cases:
the array is empty: the result is trivially 0
the first element is even: the result will be the sum of odd elements of the rest of the array
the first element is odd: the result will be this element added to the sum of odd elements of the rest of the array
In this problem the trivial case is computing the result for an empty array and the simpler version of the problem is working on a smaller array. It is important to understand that the simpler version must be "closer" to a trivial case for recursion to work.
Once the algorithm is clear translation to code is simple:
// Returns the sums of all odd numbers in
// the sequence of n elements pointed by p
int oddSum(int *p, int n) {
if (n == 0) {
// case 1
return 0;
} else if (p[0] % 2 == 0) {
// case 2
return oddSum(p + 1, n - 1);
} else {
// case 3
return p[0] + oddSum(p + 1, n - 1);
}
}
Recursion is a powerful tool to know and you should try to understand this example until it's 100% clear how it works. Try starting rewriting it from scratch (I'm not saying you should memorize it, just try rewriting it once you read and you think you understood the solution) and then try to solve small variations of this problem.
No amount of reading can compensate for writing code.
You are passing updated n to recursive function as argument but not using it inside.
change MAX to n in this statement
for(int i = 0; i < n; i ++)
so this doesnt really answer your question but it should help.
So, your code is not really recursive. If we run through your function
int total = 0; //Start a tally, good.
if (n <= 0)
return 0; //Check that we are not violating the array, good.
else
for(int i = 0; i < MAX; i ++)
{
if(arrayNum[i] % 2 != 0) //THIS PART IS WIERD
total += arrayNum[i];
}
And the reason it is wierd is because you are solving the problem right there. That for loop will run through the list and add all the odd numbers up anyway.
What you are doing by recursing could be to do this:
What is the sum of odd numbers in:
13,14,8,7,45,89,22,18,6,10
+
14,8,7,45,89,22,18,6
+
8,7,45,89,22,18
+
7,45,89,22 ... etc
And if so then you only need to change:
for(int i = 0; i < MAX; i ++)
to
for(int i = 0; i < n; i ++)
But otherwise you really need to rethink your approach to this problem.
It's not recursion if you use a loop.
It's also generally a good idea to separate computation and output.
int sum(int arrayNum[], int n)
{
if (n <= 0) // Base case: the sum of an empty array is 0.
return 0;
// Recursive case: If the first number is odd, add it to the sum of the rest of the array.
// Otherwise just return the sum of the rest of the array.
if(arrayNum[0] % 2 != 0)
return arrayNum[0] + sum(arrayNum + 1, n - 1);
else
return sum(arrayNum + 1, n - 1);
}
int main()
{
int x[MAX] = {13,14,8,7,45,89,22,18,6,10};
cout << sum(x,MAX);
}
I wrote this code in order to find the number of the divisors of a given number. The method I am trying to implement finds all of the prime factors (which works) and takes number of similar prime numbers plus one (which gives the number of divisors).
e.g. 28 = 2*2 * 7 --> (2+1)*(1+1) = 6
This is my attempt:
int num = 20;
int next = 0;
int exponent = 0;
int numberOfDivisors = 1;
start:
for (int i = 2; i <= num; i++)
{
next = i;
if (num%i == 0)
{
if (i == next)
{
exponent++;
}
else
{
numberOfDivisors *= (exponent+1);
exponent = 0;
}
if (num != i)
{
num /= i;
goto start;
}
}
}
std::cout << numberOfDivisors << std::endl;
I just cannot figure out what I am missing.
Using goto is bad as you can see from the comments. After a little bit of clean up your code comes down to:
#include <iostream>
int main() {
int num = 20;
int numberOfDivisors = 1;
for (int i = 2; i <= num; i++)
{
int exponent = 0;
while (num % i == 0) {
exponent++;
num /= i;
}
numberOfDivisors *= (exponent+1);
}
std::cout << numberOfDivisors << std::endl;
return 0;
}
The goto isn't helpful, actually, since you're already checked for lower prime factors there's no need to put i back to 2.
int numberOfDivisors = 1;
int exponent = 1;
int i = 2;
while (i <= num) {
if (num%i == 0) {
exponent++;
num /= i;
}
else {
numberOfDivisors *= exponent;
exponent = 1;
i++;
}
}
numberOfDivisors *= exponent; // <-- you were missing this, mainly
The actual problem with your code that wasn't style or performance related was that your loop finished with some exponent that you never included.
There are many issues with your code (and that goto interrupting the loop is just ugly), but one major problem with your code is that right after you enter the for loop, you set next to have the same value as i.
Then, you test something (namely, if i divides num evenly, but this doesn't matter) and (here is the critical part, you test if i is equal to next, which you just set them equal.
So, your code never has the opportunity to enter the else part of the corresponding if, which is where you would multiply the number of divisors by the exponent + 1.
As a result, your code always prints 1, which is not what you wanted.
P.S.: That else may be entirely optimized out by the compiler as part of dead-code elimination, and this is what my gcc 4.7 actually does if I enable optimization -O1 or higher.