I am stuck on what to do next... The program is suppose to check to see if entered Zero-One Matrix is an Equivalence relation (transitive, symmetric, and reflexive) or not. I am still new to C++ (started this semester). I know how to create the matrix using vector but not on how to check if it is equivalence relation or not..
I assume I need to use boolean function but I'm stuck on what I need to put in as an argument or if this is correct. My original thought was... so for symmetric it will look like (which I know this goes after #include and beofre int main(). Any help would be awesome.
bool isSymmetric(vector<int> &vect, int Value)
{
for (int i = 0; i < Value; i++)
for (int j = 0; j < Value; j++)
if (vect[i][j] != vect[j][i])
return false;
return true;
}
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector< vector<int> > vec;
cout << "NxN matrix N: ";
int Value;
cin >> Value;
cout << Value << "x" << Value << " matrix\n";
for (int i = 0; i < Value; i++) {
vector<int> row;
for (int j = 0; j < Value; j++) {
cout << "Enter a number (0 or 1): ";
int User_num;
cin >> User_num;
while (User_num != 0 && User_num != 1) {
cout << "Invalid Entry! Enter 0 or 1!\n";
cout << "Enter a number (0 or 1): ";
cin >> User_num;
}
row.push_back(User_num);
}
vec.push_back(row);
}
cout << endl;
for (int i = 0; i < Value; i++) {
for (int j = 0; j < Value; j++) {
cout << vec[i][j] << " ";
}
cout << endl;
}
cout << endl;
system("pause");
return 0;
}
Related
I am trying to get the height of these slashes to be a certain length based on input. So far, I have:
#include <iostream>
using namespace std;
int main() {
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You entered: " << n << "\n";
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= i; j++)
cout << '/' << '/';
cout << "\n";
}
}
I need it to then reverse and go back.
It prints:
//
////
//////
If the user entered 3.
It should print:
//
////
//////
////
//
Can anyone lead me in the right direction? I am new to cpp.
You can use a different kind of loop and add a bool variable to track when the program have reached "n". Then, after the program reaches "n", it sets the bool variable to true and starts to substract until i equals 0
Code below, read comments and ask if you have any further questions:
#include <iostream>
using namespace std;
int main()
{
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You have entered: " << n << "\n";
int i = 1;
bool reachedN = false; // tells if [i] has reached [n]
while (i != 0)
{
// Print required slashes
for (int j = 1; j <= i; j++)
{
cout << "//";
}
cout << '\n'; // new line
// Add until i == n, then substract
if (i == n)
{
reachedN = true;
}
if (reachedN)
{
--i;
}
else
{
++i;
}
}
}
If you enter 3, the output is the following:
This is one way to achieve that:
#include <iostream>
using namespace std;
int main() {
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You entered: " << n << "\n";
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= i; j++)
cout << '/' << '/';
cout << "\n";
}
for (int i = n - 1; i > 0; i--) {
for (int j = 1; j <= i; j++)
cout << '/' << '/';
cout << "\n";
}
}
This is a shorter solution with only two for-loops.
#include <iostream>
using namespace std;
int main()
{
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You entered: " << n << "\n";
n = n * 2 - 1;
int r = 0;
for (int j = 0; j < n; j++)
{
if (j > n / 2) r--;
else r++;
for (int i = 0; i < r; i++)
{
cout << '/' << '/';
}
cout << "\n";
}
return 0;
}
I am currently writing a task where the user will input a number and it'll output a number of "*" depending on the number. Eg if the user inputted a 5, the answer would be:
*
**
***
****
*****
This is my current code:
#include <iostream>
using namespace std;
int number;
char star = '*';
int main()
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
for (int i = 0; i < number; i++)
{
for (int j = i; j < number; j++)
{
cout << star;
}
cout << " " << endl;
}
return 0;
}
If the number 5 was inputted, this would output:
*****
****
***
**
*
How would I go about reversing the order so that it is ascending order rather than descending.
You would need to use the for loop in descending instead of ascending order.
Example:
cout << "Input a number between 1 and 10" << endl;
cin >> number;
for (int i = number -1; i >= 0; i--) {
for (int j = i; j < number; j++) {
cout << star;
}
cout << "*" << endl;
}
return 0;
}
You can do that by just reversing the order of the change of i:
#include <iostream>
using namespace std;
int number;
char star = '*';
int main()
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
//for (int i = 0; i < number; i++)
for (int i = number - 1; i >= 0; i--)
{
for (int j = i; j < number; j++)
{
cout << star;
}
cout << "*" << endl;
}
return 0;
}
(Just how to reverse the order of the change of i is shown. The output of this program is not as expected.)
My preference is using i as the number of stars to print and avoiding using gloval variables and using namespace std;:
#include <iostream>
int main()
{
int number;
char star = '*';
std::cout << "Input a number between 1 and 10" << std::endl;
std::cin >> number;
for (int i = 1; i <= number; i++)
{
for (int j = 0; j < i; j++)
{
std::cout << star;
}
std::cout << std::endl;
}
return 0;
}
Just go from number all the way down instead of up:
for (int i = number; i > 0; i--)
{
for (int j = i; j > 0; j--)
{
cout << star;
}
std::cout << '\n';
}
Instead of j<number use j<i and instead of j=i use j=0:
#include <iostream>
using namespace std;
int number;
char star = '*';
int main()
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
for (int i = 0; i < number; i++)
{
for (int j = 0; j < i; j++)
{
cout << star;
}
cout << "*" << endl;
}
return 0;
}
there are multiple ways to achieve what you want. You can insert everything inside and array and print the array backwards, insert everything inside an array (starting from last position) and then print it in natural order, you can do it only by using indexes (like i and a lot of other persons did here). I also added a little input check, and put the code in functions for clearness:
#include <iostream>
using namespace std;
int number;
char star = '*';
void version1();
void version2();
int main()
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
while(number < 1 || number > 10)
{
cout << "I SAID BETWEEN 1 AND 10, IS EASY, RIGHT?" << endl;
cin >> number;
}
version1();
version2();
return 0;
}
void version1()
{
cout <<"Version 1: "<<endl;
for (int i = 0; i < number; i++)
{
for (int j = i; j < number; j++)
{
cout << star;
}
cout << endl;
}
}
void version2()
{
cout << "Version 2: (only by using indexes)" << endl;
int cap_index = 0;
for (int i = 0; i < number; i++)
{
cap_index = i + 1;
for (int j = 0; j < cap_index; j++)
{
cout << star;
}
cout << endl;
}
}
btw this is a more serious (and simple) way to check input:
do
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
}while(number < 1 || number > 10);
In my case i put it this way:
starting from 0 print '*' up to i + 1 times, this way you have an ascending order print:
first time (from 0 to 1) print '*' -> print once
second time (from 0 to 2) print '*' -> prints twice
and so on up to the inserted number
Obviously this isn't not the best way to do this nor the most elegant, like in nearly every situation there are a lot of possibilities.
ps. i'm not sure, but you have an error in your code, as it prints a '*' more, each time, i don't know how is your output like that
By using a std::string, you can avoid the need of an explicit inner loop.
#include <iostream>
#include <string>
int main()
{
int number;
char star = '*';
std::cout << "Input a number between 1 and 10" << "\n";
std::cin >> number;
for (int i = 0; i < number; i++)
{
std::cout << std::string (i+1, star) << "\n";
}
return 0;
}
Changing the second 'for' loop condition, in your function --
From,
for (int j = i; j < number; j++)
Change it to ,
for (int j = 0; j <= i; j++)
I want to find the factors of a product by using arrays to check whether they equal it or not
the values do not print
here is the code
#include <iostream>
using namespace std;
int main() {
int arr[5] = { 1,3,5,7,2 };
int arr1[5] = { 0,6,5,4,9 };
int X;
cout << "Please enter X:"; cin >> X;
for (int i = 0, j = 0; i < 5 && j < 5; ++i, ++j) {
if (arr[i]*arr[j]==X) {
cout << arr[i] << " ";
cout << arr1[j] << " ";
}
}
}
Use this nested loops
for (int i = 0; i < 5 ;++i) {
for(int j=0 ;j<5;++j){
if (arr[i]*arr1[j]==X) {
cout << arr[i] << " ";
cout << arr1[j] << " ";
}
}
}
I am Having Problem with Passing a 2D array to a c++ Function. The function is supposed to print the value of 2D array. But getting errors.
In function void showAttributeUsage(int)
Invalid types for int(int) for array subscript.
I know the problem is with the syntax in which I am passing the particular array to function but I don't know how to have this particular problem solved.
Code:
#include <iostream>
using namespace std;
void showAttributeUsage(int);
int main()
{
int qN, aN;
cout << "Enter Number of Queries : ";
cin >> qN;
cout << "\nEnter Number of Attributes : ";
cin >> aN;
int attVal[qN][aN];
cout << "\nEnter Attribute Usage Values" << endl;
for(int n = 0; n < qN; n++) { //for looping in queries
cout << "\n\n***************** COLUMN " << n + 1 << " *******************\n\n";
for(int i = 0; i < aN; i++) { //for looping in Attributes
LOOP1:
cout << "Use(Q" << n + 1 << " , " << "A" << i + 1 << ") = ";
cin >> attVal[n][i];
cout << endl;
if((attVal[n][i] > 1) || (attVal[n][i] < 0)) {
cout << "\n\nTHE VALUE MUST BE 1 or 0 . Please Re-Enter The Values\n\n";
goto LOOP1; //if wrong input value
}
}
}
showAttributeUsage(attVal[qN][aN]);
cout << "\n\nYOUR ATTRIBUTE USAGE MATRIX IS\n\n";
getch();
return 0;
}
void showAttributeUsage(int att)
{
int n = 0, i = 0;
while(n != '\0') {
while(i != '\0') {
cout << att[n][i] << " ";
i++;
}
cout << endl;
n++;
}
}
I really suggest to use std::vector : live example
void showAttributeUsage(const std::vector<std::vector<int>>& att)
{
for (std::size_t n = 0; n != att.size(); ++n) {
for (std::size_t i = 0; i != att.size(); ++i) {
cout << att[n][i] << " ";
}
cout << endl;
}
}
And call it that way:
showAttributeUsage(attVal);
Looking at your code, I see no reason why you can't use std::vector.
First, your code uses a non-standard C++ extension, namely Variable Length Arrays (VLA). If your goal is to write standard C++ code, what you wrote is not valid standard C++.
Second, your initial attempt of passing an int is wrong, but if you were to use vector, your attempt at passing an int will look almost identical if you used vector.
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
typedef std::vector<int> IntArray;
typedef std::vector<IntArray> IntArray2D;
using namespace std;
void showAttributeUsage(const IntArray2D&);
int main()
{
int qN, aN;
cout << "Enter Number of Queries : ";
cin >> qN;
cout << "\nEnter Number of Attributes : ";
cin >> aN;
IntArray2D attVal(qN, IntArray(aN));
//... Input left out ...
showAttributeUsage(attVal);
return 0;
}
void showAttributeUsage(const IntArray2D& att)
{
for_each(att.begin(), att.end(),
[](const IntArray& ia) {std::copy(ia.begin(), ia.end(), ostream_iterator<int>(cout, " ")); cout << endl;});
}
I left out the input part of the code. The vector uses [] just like a regular array, so no code has to be rewritten once you declare the vector. You can use the code given to you in the other answer by molbdnilo for inputing the data (without using the goto).
Second, just to throw it into the mix, the showAttributeUsage function uses the copy algorithm to output the information. The for_each goes throw each row of the vector, calling std::copy for the row of elements. If you are using a C++11 compliant compiler, the above should compile.
You should declare the function like this.
void array_function(int m, int n, float a[m][n])
{
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
a[i][j] = 0.0;
}
where you pass in the dimensions of array.
This question has already been answered here. You need to use pointers or templates. Other solutions exists too.
In short do something like this:
template <size_t rows, size_t cols>
void showAttributeUsage(int (&array)[rows][cols])
{
for (size_t i = 0; i < rows; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < cols; ++j)
std::cout << array[i][j] << '\t';
std::cout << std::endl;
}
}
You're using a compiler extension that lets you declare arrays with a size determined at runtime.
There is no way to pass a 2D array with such dimensions to a function, since all but one dimension for an array as a function parameter must be known at compile time.
You can use fixed dimensions and use the values read as limits that you pass to the function:
const int max_queries = 100;
const int max_attributes = 100;
void showAttributeUsage(int array[max_queries][max_attributes], int queries, int attributes);
int main()
{
int attVal[max_queries][max_attributes];
int qN = 0;
int aN = 0;
cout << "Enter Number of Queries (<= 100) : ";
cin >> qN;
cout << "\nEnter Number of Attributes (<= 100) : ";
cin >> aN;
cout << "\nEnter Attribute Usage Values" << endl;
for (int n = 0; n < qN; n++)
{
cout << "\n\n***************** COLUMN " << n + 1 <<" *******************\n\n";
for (int i = 0; i < aN; i++)
{
bool bad_input = true;
while (bad_input)
{
bad_input = false; // Assume that input will be correct this time.
cout << "Use(Q" << n + 1 << " , " << "A" << i + 1 << ") = ";
cin >> attVal[n][i];
cout << endl;
if (attVal[n][i] > 1 || attVal[n][i] < 0)
{
cout << "\n\nTHE VALUE MUST BE 1 or 0 . Please Re-Enter The Values\n\n";
bad_input = true;
}
}
}
}
cout << "\n\nYOUR ATTRIBUTE USAGE MATRIX IS\n\n";
showAttributeUsage(attVal, qN, aN);
getch();
return 0;
}
void showAttributeUsage(int att[max_queries][max_attributes], int queries, int attributes)
{
for (int i = 0; i < queries; i++)
{
for (int j = 0; j < attributes; j++)
{
cout << att[i][j] << " ";
}
cout << endl;
}
}
For comparison, the same program using std::vector, which is almost identical but with no size limitations:
void showAttributeUsage(vector<vector<int> > att);
int main()
{
cout << "Enter Number of Queries (<= 100) : ";
cin >> qN;
cout << "\nEnter Number of Attributes (<= 100) : ";
cin >> aN;
vector<vector<int> > attVal(qN, vector<int>(aN));
cout << "\nEnter Attribute Usage Values"<<endl;
for (int n = 0; n < qN; n++)
{
cout<<"\n\n***************** COLUMN "<<n+1<<" *******************\n\n";
for (int i = 0; i < aN; i++)
{
bool bad = true;
while (bad)
{
bad = false;
cout << "Use(Q" << n + 1 << " , " << "A" << i + 1 << ") = ";
cin >> attVal[n][i];
cout << endl;
if (attVal[n][i] > 1 || attVal[n][i] < 0)
{
cout << "\n\nTHE VALUE MUST BE 1 or 0 . Please Re-Enter The Values\n\n";
bad = true;
}
}
}
}
cout << "\n\nYOUR ATTRIBUTE USAGE MATRIX IS\n\n";
showAttributeUsage(attVal);
getch();
return 0;
}
void showAttributeUsage(vector<vector<int> > att);
{
for (int i = 0; i < att.size(); i++)
{
for (int j = 0; j < att[i].size(); j++)
{
cout << att[i][j] << " ";
}
cout << endl;
}
}
The Particular Logic worked for me. At last found it. :-)
int** create2dArray(int rows, int cols) {
int** array = new int*[rows];
for (int row=0; row<rows; row++) {
array[row] = new int[cols];
}
return array;
}
void delete2dArray(int **ar, int rows, int cols) {
for (int row=0; row<rows; row++) {
delete [] ar[row];
}
delete [] ar;
}
void loadDefault(int **ar, int rows, int cols) {
int a = 0;
for (int row=0; row<rows; row++) {
for (int col=0; col<cols; col++) {
ar[row][col] = a++;
}
}
}
void print(int **ar, int rows, int cols) {
for (int row=0; row<rows; row++) {
for (int col=0; col<cols; col++) {
cout << " | " << ar[row][col];
}
cout << " | " << endl;
}
}
int main () {
int rows = 0;
int cols = 0;
cout<<"ENTER NUMBER OF ROWS:\t";cin>>rows;
cout<<"\nENTER NUMBER OF COLUMNS:\t";cin>>cols;
cout<<"\n\n";
int** a = create2dArray(rows, cols);
loadDefault(a, rows, cols);
print(a, rows, cols);
delete2dArray(a, rows, cols);
getch();
return 0;
}
if its c++ then you can use a templete that would work with any number of dimensions
template<typename T>
void func(T& v)
{
// code here
}
int main()
{
int arr[][7] = {
{1,2,3,4,5,6,7},
{1,2,3,4,5,6,7}
};
func(arr);
char triplestring[][2][5] = {
{
"str1",
"str2"
},
{
"str3",
"str4"
}
};
func(triplestring);
return 0;
}
I need help with getting this users input of an integer and retrieving the even numbers and displaying them with spaces.I already have the input processed into an array and have it reversed (thanks to stackoverflow) now need to extract the even numbers from the array and display them.
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
int evenNumbers(char even[], int num[], int indexing[]);
int main()
{
char integers[5];
int numbers[5];
int even[5] = {0,2,4,6,8};
int evens;
cout << "Please enter an integer and press <ENTER>: " << endl;
for (int j = 0; j < 5; j++)
cin >> integers[j];
for (int j = 0; j < 5; j++)
{
numbers[j]= integers[j] - '0';
}
cout << endl;
for (int j = 5; j > 0; j--)
{
cout << integers[j - 1] << " ";
}
cout << endl;
//having problems finding the even numbers and displaying the even numbers
//from the users input of integers, i have only learned how to display the
//subscript by a linear search
evens = evenNumbers(integers, numbers, even);
if (evens == -1)
cout << "There are no even numbers" << endl;
else
{
cout << "The even numbers are: " << (evens + 1) << endl;
}
system("pause");
return 0;
}
int evenNumbers(char even[], int num[], int indexing[])
{
int index = 0;
int position = -1;
bool found = false;
for (int j = 0; j < 5; j++)
{
num[j]= even[j] - '0';
}
while (index < 5)
{
if (num[index] == indexing[index])
{
found = true;
position = index;
}
index++;
}
return position;
}
If you want to display the even numbers from the array integers you can use a simple for loop and if statement:
for(int i = 4; i >= 0; i--)
{
if(integers[i] % 2 == 0)
cout << integers[i] << " ";
}
Your approach is all wrong, you can't detect even numbers by searching a list, you need a mathematical test for evenness. Write a function called is_even which tests one number and returns true if it is even and false if it is not. Then you can use that function, very simply, like this
for (int j = 0; j < 5; j++)
{
if (is_even(integers[j]))
cout << integers[j] << " ";
}
cout << endl;
Now you just need to write the is_even function.
void evennumbers(int num[])
{
for(int i=0;i<5;i++)
{
if(num[i]%2==0)
cout<<num[i]<<" ";
}
}
And avoid taking input to char what if user enters a number with more than one digit
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
void validNum(char valid[]);
void reverseNum(char rev[], int num2[]);
void evenNumbers(char even[], int num3[]);
void oddNumbers(char odd[], int num4[]);
int main()
{
char integer[5];
int number[5];
cout << "Your number is: ";
validNum(integer);
cout << "Your number in reverse is: ";
reverseNum(integer, number);
cout << "Even numbers: ";
evenNumbers(integer, number);
cout << endl;
cout << "Odd numbers: ";
oddNumbers(integer, number);
cout << endl;
system("pause");
return 0;
}
void validNum(char valid[])
{
char ch;
cout << "Please enter an integer and press <ENTER>: " << endl;
ch = cin.get;
while (ch < 0 || ch >= 'A' && ch <= 'Z' || ch >= 'a' && ch <= 'z')
{
cout << "ERROR: Please enter a positive integer and press <ENTER>: ";
for (int i = 0; i < 5; i++)
cin >> valid[i];
}
for (int j = 0; j < 5; j++)
{
cout << valid[j] - '0';
}
}
void reverseNum(char rev[], int num2[])
{
for (int j = 0; j < 5; j++)
{
num2[j]= rev[j] - '0';
}
cout << endl;
for (int j = 5; j > 0; j--)
{
cout << rev[j - 1]<< " ";
}
cout << endl;
}
void evenNumbers(char even[], int num3[])
{
for (int i = 0; i < 5; i++)
{
if (even[i] % 2 == 0)
{
cout << num3[i] << " ";
}
}
}
void oddNumbers(char odd[], int num4[])
{
for (int i = 0; i < 5; i++)
{
if (odd[i] % 2 == 1)
{
cout << num4[i] << " ";
}
}
}