A={O|0<x<=10}
B={E|0<x<=10}
R={(a,b)a A, b B|a<b}
A={2,4,6,8,10}
B={1,3,5,7,9}
AxB={(2,1)(2,3)(2,5)(2,7)(2,9) (4,1)(4,3)(4,5)(4,7)(4,9) (6,1)(6,3)(6,5)(6,7)(6,9) (8,1)(8,3)(8,5)(8,7)(8,9) (10,1)(10,3)(10,5)(10,7)(10,9)}
R={(2,3)(2,5)(2,7)(2,9) )(4,5)(4,7)(4,9) )(6,7)(6,9) (8,9)}
Solve it using C++ program. You can use Arrays for storing the values and use nested for loop for finding the R(Relation). The above code is a Desecrate mathematics question and i want to solve it using C++. i want to search from the arrays i write this code kindly guide me what is wrong in my code...
#include<iostream>
using namespace std;
main()
{
int a[5]= {2,4,6,8,10}, b[5]={1,3,5,7,9} ,R1[10],R2[10], counts =0;
for (int i=0; i<5; ++i)
{
for (int j=0; j<5; j++)
{
if (a[i]<b[j])
{
R2[counts]= b[j];
R1[counts]= a[i];
counts++;
}
}
}
for(int i=0; i<counts; i++)
{
cout<<R2[i]<<endl;
cout<<R1[i];
}
}
for(int i=0; i<counts; i++)
{
cout << "(" << R1[i] << "," << R2[i] << ")" << endl;
}
Related
My code works when put in the int main() function but when I implement it as another function (void bubbleSort) the output displays it as if there was no sorting done.
void bubbleSort(int numeros[])
{
int store = 0;
int length = ARRAY_SIZE(numeros);
for(int i=0; i<(length-1); i++)
{
for(int j=0; j<(length-i-1); j++)
{
if(numeros[j] < numeros[j+1])
{
store = numeros[j];
numeros[j] = numeros[j+1];
numeros[j+1] = store;
}
}
}
for(int m=0; m<1000; m++)
{
cout << numeros[m] <<' ';
}
}
What could I have possibly done wrong? Any help would be greatly appreciated.
You can't pass a full array as an argument to a c++ function, only a pointer to the first element in the array. As a result you need some way to tell the function how long the array is. One way it to pass that in as another argument (as shown below). There is some discussion and suggestions of other/better ways to do it here.
For example if you accidentally pass in the wrong length argument to these functions they will start operating on whatever memory exists after the block of memory where your array is.
#include <iostream>
using namespace std;
void printArray(int array[], int length) {
for(int i=0; i<length; i++) {
cout << array[i] << " ";
}
cout << endl;
}
void bubbleSort(int numeros[], int length) {
int store = 0;
for(int i=0; i<(length-1); i++) {
for(int j=0; j<(length-i-1); j++) {
if(numeros[j] < numeros[j+1]) {
store = numeros[j];
numeros[j] = numeros[j+1];
numeros[j+1] = store;
}
}
}
cout << "array at end of bubble sort: ";
printArray(numeros, length);
}
int main() {
int anArray[] = {1, 3, 2, 4, 6, 5, 10, 9, 7, 8};
int arraySize = sizeof(anArray)/sizeof(anArray[0]);
cout << "arraySize: " << arraySize << endl;
cout << "array before sort: ";
printArray(anArray, arraySize);
bubbleSort(anArray, arraySize);
cout << "array after sort: ";
printArray(anArray, arraySize);
return 0;
}
I try to enter 2d array and to sum all numbers in one row. Then I convert that number to binary (8 bit) and to set it again in new 2d array. Here's my code. I get output in negative numbers and I expect binary number.
I input
1 2 3
4 5 6
7 8 9
And i want this output
00000110
00001111
00011000
i get
00000000
00000000
00000000
#include<iostream>
using namespace std;
int main()
{
int n,m,j,i;
int a[50][50],b[50][8],c[50];
cin>>n>>m;
for(i=0;i<n;i++)
{
c[i]=0;
for(j=0;j<m;j++)
{
cin>>a[i][m];
cin.ignore();
c[i]+=a[i][j];
}
}
for(i=0;i<n;i++)
for(j=0;j<8;j++)
{
b[i][j]=c[i]%2;
c[i]/=2;
}
for(i=0;i<n;i++)
{
for(j=0;j<8;j++)
{
cout<<b[i][j];
}
cout<<endl;
}
}
I attempted to revise your code but I soon realised that you were doing some weird unnecessary things so I just started fresh and here's what I've got for you:
#include <iostream>
#include <vector>
using namespace std;
void add_array(int arr1[], int arr2[], int arrLength, int ret[]) {
for(int i = 0; i < arrLength; i++) {
ret[i] = arr1[i]+arr2[i];
}
return;
}
void to_binary(int n, vector<int> *ret) {
while(n!=0) {
ret->push_back(n%2==0 ?0:1);
n/=2;
}
}
int main() {
int a[5] = {1,2,3,4,5};
int b[5] = {6,7,8,9,10};
int c[5];
add_array(a, b, 5, c);
cout << "A:" << endl;
for(int i = 0; i < 5; i++) {
cout << i << " : " << a[i] << endl;
}
cout << "B:" << endl;
for(int i = 0; i < 5; i++) {
cout << i << " : " << b[i] << endl;
}
cout << "C:" << endl;
for(int i = 0; i < 5; i++) {
cout << i << " : " << c[i] << endl;
}
vector<int> vec;
for(int i = 0; i < 5; i++) {
to_binary(c[i], &vec);
for(int j = 0; j < vec.size(); j++) {
cout << vec[j];
}
cout << endl;
vec.clear();
}
return 0;
}
I don't know how you were handling the adding of the two functions so I just wrote a really simple function there, I'll start with the parameters
int arr1[], int arr2[]
These are the two functions you'll be adding, simple.
int arrLength
This tells the function what the length of the two arrays is for the 'for loop'
int ret[]
Ret is the return array and is passed in so that it can be modified with the added arrays, now you could do that with either of the other two arrays but this is better practice especially if you want to reuse the other arrays later.
Now here's the function itself
for(int i=0;i<arrLength;i++){ ret[i]=arr1[i]+arr2[i];}
Here we loop through each position in the arrays and place them in the variable 'ret', this is the whole thing.
The function
void to_binary(int n, vector<int> *ret)
handles the decimal to binary using a vector for variable sizes, it's basically what you were doing just in a function.
In the function main we create the three arrays and call add_array with the necessary arguments, then we create the vector vec and then proceed to loop through the array c getting the binary number of each position and then since we stored the binary number in an int vector instead of a string we loop through the vector
for(int j = 0; j < vector.size(); j++)
We are using vector.size() to get the dynamic size of the vector, then we print out each binary digit and then print an endl and clear the vector for reuse.
I need to create 2 arrays, each with 4 elements. One array contains four int values gotten from the user, and the other array contains pointers to the elements of the first array. I keep getting the following error:
array type 'int *[4]' is not assignable
on this line:
my_ptrs = &my_ints;
Here is my code:
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int my_ints[4];
int *my_ptrs[4];
float temp;
int num;
for (int x=0; x< 4; x++)
{
cout << "Enter Integer:" << endl;
cin >> num;
my_ints[x] = num;
}
my_ptrs = &my_ints;
for(int k=0; k<=3; k++)
{
for(int j=k+1; j<=3; j++)
{
if(my_ptrs[k]>my_ptrs[j])
{
temp=*my_ptrs[k];
my_ptrs[k]=my_ptrs[j];
*my_ptrs[j]=temp;
}
}
cout << my_ptrs[k] << " ";
}
return 0;
}
Your apparent intent is to have each pointer in my_ptrs to point to the corresponding value in my_ints.
I'm afraid there are no shortcuts here, using a single assignment. You have to do it the hard way:
for (int i=0; i<4; ++i)
my_ptrs[i]=&my_ints[i];
You cannot assign a pointer-to-an-array to an array variable, like you are trying to do. But what you can do instead is either:
initialize the second array in its declaration directly:
int my_ints[4];
int* my_ptrs[4] = {&my_ints[0], &my_ints[1], &my_ints[2], &my_ints[3]};
populate the array in a separate loop:
int my_ints[4];
int* my_ptrs[4];
for (int i = 0; i < 4; ++i)
my_ptrs[i] = &my_ints[i];
That being said, based on what you are trying to do ("i want to print the my_ptrs array in ascending order"), you could just get rid of the second array altogether and use the std::sort() algorithm instead:
Sorts the elements in the range [first, last) in ascending order.
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int my_ints[4];
for (int x = 0; x < 4; ++x)
{
cout << "Enter Integer:" << endl;
cin >> my_ints[x];
}
std::sort(my_ints, my_ints+4);
for(int k = 0; k < 4; ++k)
cout << my_ints[k] << " ";
return 0;
}
I am trying to pass c-strings into a function and have a table 6x6 printed out. However my code keeps compiling with an error. A segmentation fault. I understand that a seg fault is when the program looks for something that isn't there. I'm not sure how to fix this or how to alter this even after much research. Any suggestions would very much be appreciated.
Here's my code:
#include <iostream>
using namespace std;
void fillarrays(int labs[][6]);
void printarrays(int labs[][6]);
int main()
{
int labs[6][6];
fillarrays(labs);
printarrays(labs);
return 0;
}
void fillarrays(int labs[][6])
{
for(int row = 0; row<=6; row ++)
{
for(int col=0; col<=6; col++)
{
labs[row][col] = row;
}
}
}
void printarrays(int labs[][6])
{
cout << "Labs: " << '\t' << '\t' << endl;;
for(int rows=0 ; rows<6; rows++)
{
for(int cols=0; cols<6; cols++)
{
cout << '\t' << labs[rows][cols] << " ";
}
cout << endl;
}
}
labs[6][6] is an array with 6 rows first one is [0] and last one is [5], the same for cols, so change <=6 to <6 in fillarrays.
for(int row = 0; row<6;row ++) {
for(int col=0; col<6; col++) {
labs[row][col] = row;
}
}
for(int row = 0; row<=6;row ++) {
for(int col=0; col<=6; col++) {
labs[row][col] = row;
}
}
Look closely here - this fills a 7x7 array (indices 0 to 6 for rows and columns). But the array is only 6x6, so you're trying to write outside teh bounds of the array.
This is a reasonably common mistake; change the <=s to <s.
So I wrote a program that is supposed select the perfect squares from an array and put it into another array. Example: (2,4,13,5,25,66) and the second array(the result) should look like this (4,25)
My result looks like this (0,4,0,0,25,0) ...so its half good ...how to make it show only 4,25 ?
#include<iostream.h>
#include<math.h.>
int main()
{
int A[100],i,n,p,j;
cout << "Number of array elements=";
cin >> n;
for(i=1;i<=n;i++)
{
cout<<"A["<<i<<"]=";
cin>>A[i];
}
for(i=1;i<=n;i++)
{
p=sqrt(A[i]) ;
if(p*p==A[i])
A[j]=A[i];
else
A[i]=0;
cout << A[i] << " ";
}
return 0;
}
USING ONLY c++ basic commands...as i did
You need to keep a separate count of how many perfect squares you've found and use that to place your answers into an array of perfect squares:
int squares[???];
// ...
if(p*p==A[i]) {
squares[squaresFound]=A[i];
squaresFound++;
}
The problem now will be to decide how long the squares array should be. You don't know ahead of time how many squares you're going to get. Are you going to have it the same size as A and fill the rest with 0s? Or do you want the array of squares to be exactly the right size?
If you want it to be the right size, you're much better off using a std::vector:
std::vector<int> squares;
// ...
if(p*p==A[i]) {
squares.push_back(A[i]);
}
But I think your silly "only basic C++ commands" restriction will not allow you to do this.
You talk about a second array (the result), yet your code declares only one array! Additionally, you reference A[j], but your j has not be initialized.
You should declare another array B[100], initialize j to zero, and then use this code when you find a square:
int j = 0;
for (int i=0 ; i != n ; i++) {
int p = sqrt(A[i]);
if(p*p==A[i]) {
B[j++] = A[i];
}
}
Make another array, remove all occurrences of 0 from the resultArray and add non-0 to newArray.
OR
int j=0
if(A[i]==p*p)
squares[j++]=A[i];
#include <cmath>
#include <iostream>
using namespace std;
int main()
{
int A[100];
int n;
cout << "Number of array elements = " << endl;
cin >> n;
for(int i = 0; i < n; i++)
{
cout << "A[" << i << "] = ";
cin >> A[i];
}
int B[100];
int cnt_sqr = 0;
for(int i = 0; i < n; i++)
{
int p = sqrt(A[i]);
if (p * p == A[i])
{
B[cnt_sqr++] = A[i];
}
}
for (int i = 0; i < cnt_sqr; i++)
{
cout << B[i] << ' ';
}
return 0;
}
Full code of that about what you were told above
If you do not want to modify your code you can write the following:
for(i=1;i<=n;i++)
{
p=sqrt(A[i]) ;
if(p*p==A[i])
{
cout << A[i] << " ";
}
}
It will print you only perfect squares.
If you want to copy elements to another array:
int squares[100] = {0}; // Assuming that all values can be perfect squares
int square_count = 0;
for(i=1;i<=n;i++)
{
p=sqrt(A[i]) ;
if(p*p==A[i])
{
squares[square_count++] = A[i];
}
}