Hello taking an online class on nested loops and this was provided as the example but I don't really know what is going on.
The following code example shows nesting for loops to output a chess or checkerboard representation using the characters X and O. Why do we need x and y variables to execute a certain amount of times. And what does alternate = !alternate; mean? About the x and y wouldn't it just do it 8 times total because its greater than the amount of times y supplies? what is the difference in purpose for the two for statements? Thank you.
for (int x = 0; x < 8; x++)
{
for (int y = 0; y < 4; y++)
{
if (alternate)
{
cout << "X ";
cout << "O ";
}
else
{
cout << "O ";
cout << "X ";
}
}
alternate = !alternate;
cout << endl;
}
The variable x used for the number of lines you want to print X-O pairs. Variable y used to specify the number of X-O pairs in 1 line. So for printing 8 lines of X-O pairs and in each line, 4 pairs of X-O, you should do just like that.
The operator '!' used for getting the opposite of a value (it's logical NOT) (for example, 1 to 0 or false to true). so alternate = !alternate; means that after every line of X-O pairs, it changes from true to false or vise versa.
So lines' first character (X or O) will change according to 'alternate' variable.
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Tried several times yet still didnt manage to find my mistake: here is my program. i need to find the odd numbers from 1 and integer x and find the sum of them cubed.
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int x;
int i = 1;
int result;
cout <<" Enter the value for n" << endl;
cin >> x;
while (i >x)
if (i%2 == 0) {}
else {
result += pow(i,3);
i++;
}
cout << "The sum of odd integers cubes from " << i << " to " << x << "= " << result << endl;
return 0;
}
Minimally, you should change the compare in while
from
while (i > n)
to
while (i <= n)
There a many numbers where i will be greater than the number entered, n.
You didn't add in your curly brackets for the while loop.
while (i > x)
if (i%2 == 0) {}
needs to be:
while (i > x){
if (i % 2 == 0) {}
}
Plus what are you doing inside of that if statement? You should decrement x, to find if each number is odd.
Plus, your program ends early because i is 1 and if the user enters a number above 1, your while loop won't even run. You're telling the while loop to run ONLY when i is larger than x. Try changing it to less than:
from:
while (i > x){
to:
while (i < x){
Plus you're not doing anything with x. You want to decrement x, not add i. Although, I would recommend using a do-while loop. ( a dowhile loop does one iteration first before incrementation)
do{
if (x % 2 == 0) { // if the remainder of x/2 is 0, do this
x--;
cout << "Equal: " << x << endl;
}
if(x % 2 != 0) { //if the remainder of x/2 is not 0, do this.
temp = pow(x,3);
//you don't want to take the power of the added sum,
//you were taking the power 3 of x which was being added to.
//you want to add the sums of each power. So here we have temp, a
//temporary variable to store your addends.
result = result + temp;
cout << "Not equal, temp:" << temp <<endl;
cout << "Result: "<< result << endl;
x--; //you didn't have a decrement, you need to bring x eventually down to i if you want the loop to end, or even look through all of the numbers
}
}
while (i < x);
//You have to have this semi colon here for the compiler to know its a do-while.
cout << "The sum of odd integers cubes from " << i << " to " << userVar
<< " = " << result << endl;
return 0;
}
note: if-else statements are for flow control, its like true and false, one or the other, so that your data will flow somewhere. I used two if statements because I want to have complete control over the flow.
note2: It's ok to use:
using namespace std;
at first, but eventually you want to start learning what library each command is using. When you get into more complex programming, you start using commands from different libraries than the standard one.
I am trying to create a program to print first 200 elements following a specific numerical series condition which is
1-1-3-6-8-8-10-20
But instead of showing, just 200 elements is showing 802. I assume is because of the code inside the for loop. I have hours thinking on how to reduce that code to the job and I cannot think anything else. I am getting frustrated and need your help.
The exercise is on the code comments
//Print the following numerical series 1-1-3-6-8-8-10-20 until 200
#include <stdafx.h>
#include <iostream>
#include <stdlib.h>
using namespace std;
int main()
{
int Num1=200, z = 0, x = 1, y = 1;
cout << "\n\n1,";
cout << " 1,";
for (int i = 1; i <= Num1; i++)
{
z = y + 2;
cout << " " << z << ","; //It will print 3
z = z * 2;
cout << " " << z << ",";//It will print 6
z = z + 2;
cout << " " << z << ",";//It will print 8
z = z;
cout << " " << z << ",";//It will print 8
y = z;
}
cout << "\n\n";
system("pause");
return 0;
}
You're looping 200 times, and each time you loop, you're printing out 4 different numbers. You're also printing twice at the start so thats 2 + 4 * 200 = 802, which is where your 802 number output is coming from.
I assume is because of the code inside the "for" loop but I've hours
thinking on how to reduce that code to the job and I cannot think
anything else. I'm getting frustrated and need your help.
So you basically wanna simplify your code. Which can be done by noticing the repetitions.
There you can find only two types of change in the series; either a +2 or x2 with the previous element.
In each iteration this can be achieved by:
If reminder i%4 == 1 or i%4 == 3, need an increment of 2 (assuming 1 <= i <= MAX)
If reminder i%4 == 0, nothing but a multiplication of 2.
When you do like so, you can simply neglect, printing of first two ones and other complications in the total numbers in the series.
Also not that, you are trying to get 200 terms of this series, which increases in each step very fast and exceed the maximum limit of int. Therefore, long long is needed to be used instead.
The updated code will look like this:
#include <iostream>
typedef long long int int64;
int main()
{
int size = 200;
int64 z = -1;
for (int i = 1; i <= size; i++)
{
if ((i % 4 == 1) || (i % 4 == 3)) z += 2;
else if (i % 4 == 0) z *= 2;
std::cout << z << "\n";
}
return 0;
}
See the Output here: https://www.ideone.com/JiWB8W
I'm confused at this example:
int x = 5;
if (x==5) cout << x; // output 5
if (x==6) cout << x;
if (x=6) cout << x; // output 6
x = 0;
if (x=0) cout << x;
x = 5;
if (x-5) cout << x;
if (x-6) cout << x; // output 5
I understand first if (x==5), but why does it output 6 at if (x=6) when x = 5, and why won't it output 0 in if(x=0)
if (x=6)
means not comparison, but assignment. You assign 6 to x and the return value of the expression is 6, which is not 0 so it gains true.
similar with if (x=0) The expression x=0 gains 0 so it means if(0)
The thing about computers is that they're extraordinarily literal. A missing semicolon, or an added character, can completely change a program's function. So you need to be just as careful as a computer when working wth programs.
As #juanchopanza alluded to, there is a difference between == and = - and you already know what it is.
I have a problem that is asking for me to write a C++ program using for loops with less than 3 “cout” statements in your code to print the following pattern (ignore the pipes, the asterisks wouldn't appear without them):
|*
|***
|*****
|*******
|*********
|*********
|*******
|*****
|***
|*
This is my code I used for a fibonacci generator and I feel like it might be similar. I am able to print the "*" symbol but not in horizontal lines. What I need most help with is reversing the output. As in if given number n, I want the series to go n numbers into the series and then back down to 0.
#include <iostream>
using namespace std;
int main()
{
int y = 1, sum = 1, n;
cout << "Enter the number of terms you want" << endl;
cin >> n;
cout << "First " << n << " terms are :- " << endl;
for (int x = 0; x < n; x++) {
cout << "\n" <<endl;
for (int i = 0; i < sum; i++) {
cout << "*" << endl;
}
sum = y + 2;
y = sum;
}
}
It seems this is a homework, so I give some hints instead of a full solution.
For printing the *s in one line, please note that << endl will end the line in the output, i.e. print a line break. (The same does << "\n" by the way.) Not every cout statement has to have an << endl at its end.
For reversing the fibonacci sequence, once you have the last number in the variable sum, just do the reverse computation (i.e. subtraction). This could be done in a second set of loops, however, since you should not use cout statements too often, you better reuse the same loop by using some additional variable holding the current state (i.e. if you are counting up or down) and using an if to decide which computation to do. (I read the requirements such that only the cout statements for printing the pattern count to the "less than three" = 2)
So I create and initialize a vector (of size nmask+3) to 0, and I assign an initial value to one of the elements. I then make a for loop that goes through the first nmask elements of the vector and assigns to each element an average of 26 other elements in the vector (defined by the 4D int array voxt, which contains vector addresses).
My problem is that when I check the values of nonzero elements in my vector (phi) within the nested loop (the first cout), the values are fine and what I expect. However, when the loop finishes going through all nmask elements (for (int i= 0; i<nmask; i++) exits), I check the nonzero elements of phi again, and they are all lost (reset to 0) except for the last non-zero element (and element tvox which is manually set to 1).
I feel that since phi is initialized outside of all the loops, there should be no resetting of values going on, and that any updated elements within the nested loop should remain updated upon exit of the loop. Any ideas as to what is going on / how to fix this? Code is below; I tried to comment in a sense of the outputs I'm getting. Thanks in advance.
vector<double> phi(nmask+3, 0); //vector with nmask+3 elements all set to 0 (nmask = 13622)
phi[tvox]= 1; //tvox is predefined address (7666)
for (int n= 0; n<1; n++)
{
vector<double> tempPhi(phi); //copy phi to tempPhi
for (int i= 0; i<nmask; i++)
{
for (int a= -1; a<=1; a++)
{
for (int b= -1; b<=1; b++)
{
for (int c= -1; c<=1; c++)
{
if (!(a==0 && b==0 && c==0))
{
//oneD26 is just (double) 1/26
phi[i]= tempPhi[i]+oneD26*tempPhi[voxt[i][1+a][1+b][1+c]];
if (phi[i]!=0)
{
//this gives expected results: 27 nonzero elements (including tvox)
cout << n << " " << i << " " << a << b << c << " " << phi[i] << endl;
}
}
}
}
}
}
phi[svox]= 0; //svox = 7681
phi[tvox]= 1;
for (int q= 0; q<nmask; q++)
{
//this gives only 2 nonzero values: phi[tvox] and phi[9642], which was the last nonzero value from 1st cout
if (phi[q]!=0)
cout << q << " " << phi[q] << endl;
}
}
Difficult to tell just what is going on, but the easiest explanation is that after phi[i] gets set to non-zero and displayed to cout, it gets set to zero again in one of the later iterations through the inner loops.
If you do some tracing and check phi[i] just before updating you'll see that you often overwrite a non-zero element with zero.
Note: I have no idea what your code does, this is pure Sherlock Holmes reasoning.. if after the loops you find only 2 non-zero elements then the only logical consequence is that after updating something to non-zero later in the loop you update it to zero.
phi[i]= tempPhi[i]+oneD26*tempPhi[voxt[i][1+a][1+b][1+c]];
The nested for-loops using a, b, and c run for a combined 9 iterations with the same value of i. Since you overwrite phi[i] to a new value every time, you only retain the value from the last iteration where a, and c are all 1. If that last iteration happens to produce zero values, then phi[i] will have lots of zeroes. Perhaps you meant to do something like phi[i] += ... instead of phi[i] = ...?
I do suggest to replace the meat of the loop with something like
const boost::irange domain(-1,2);
for (int i: boost::irange(0, nmask)) for (int a: domain) for (int b: domain) for (int c: domain)
{
if (a==0 && b==0 && c==0)
continue;
//oneD26 is just (double) 1/26
phi[i]= tempPhi[i]+oneD26*tempPhi[voxt[i][1+a][1+b][1+c]];
if (phi[i]!=0)
{
//this gives expected results: 27 nonzero elements (including tvox)
cout << n << " " << i << " " << a << b << c << " " << phi[i] << endl;
}
}
Of course, for brevity I assume both boost/range.hpp and c++0x compiler. However, with trivial macro's you can achieve the same. That is without writing/using a proper combinations algorithm (why is that not in the standard, anyway).