I'm trying to create a program in c++ in which takes the users input and displays that specific row of Pascal's triangle.
i.e. inputting 3 should result in 1 3 3 1.
However when I input 3 in my program it displays "1 1 1 1". I think I'm fairly close but I can't quite work out what's wrong.
#include <iostream>
using namespace std;
int n, k, fact1, fact2, fact3, number1, number2, number3, nTakeAwayK, nChooseK;
void CalculateNChooseK();
int main()
{
cout << "Enter a row number for Pascal's Triangle: ";
cin >> n;
int x = 1;
for (int k = 0; k <= n; k++)
{
cout << x << '\t';
CalculateNChooseK();
x = nChooseK;
}
cout << endl;
system("PAUSE");
return 0;
}
void CalculateNChooseK()
{
fact1 = 1;
for (number1 = 1; number1 <= n; number1++)
{
fact1 = (fact1 * number1);
}
fact2 = 1;
for (number2 = 1; number2 <= k; number2++)
{
fact2 = (fact2 * number2);
}
nTakeAwayK = (n - k);
fact3 = 1;
for (number3 = 1; number3 <= nTakeAwayK; number3++)
{
fact3 = (fact3 * number3);
}
nChooseK = fact1 / (fact2*fact3);
}
Modify your code to this.. Hope this will help
for (k = 1; k <= n+1; k++)
{
cout << x << '\t';
CalculateNChooseK();
x = nChooseK;
}
As you are declaring k in for loop, This value is not working as global.
for (int k = 0; k <= n; k++) // don't declare k here.. and also loop from 1 to n+1 inclusive
Related
I need to do this exercise, with struct and include the function in the struct to do the calculations, but for some reason, it's not working.
I recommended you to check the image for a better idea.
#include <iostream>
using namespace std;
struct Alfa{
double h,x,n,a;
void shuma(){
cout << "enter n: "; cin >> n;
cout << "enter a: "; cin >> a;
for (int i = 1; i >= n; i++){
x = 2 * i + a;
}
};
};
int main() {
Alfa alf;
alf.shuma();
alf.h = (alf.x / 2) + 3;
cout << alf.h;
return 0;
}
You are not following the formula in the image. Use this instead:
double sum = 0;
for (int i = 1; i <= n + 1; i++) {
if (i != 4) {
sum += 2 * i + a;
}
}
h = x / 2 + 3 * sum;
for (int i = 1; i >= n; i++) means i initiated to 1, while i>=n do the loop, then inc i. So when n is bigger then 1, it will never enter the loop. Maybe you want for (int i = 1; i <= n+1; i++) ?
I need help, I created a short little program a while ago where it would print a simple pyramid with "*" like this:
*
***
*****
but I decided to challenge myself and see if I could create a simple diamond shape like this:
*
***
*****
***
*
Here is my code so far.
I should also add that the value you input, for example 5, determines how big the diamond is.
#include <iostream>
#include <sstream>
using namespace std;
int main() {
int value = 0;
cout << "Please enter in a value: ";
cin >> value;
cout << endl;
for (int i = 0; i < value; i++) {
//print spaces v v v
for (int x = 0; x < (value - i - 1); x++) {
cout << " ";
}
//print * v v v
for (int y = 0; y < (2 * i + 1); y++) {
cout << "*";
}
cout << endl;
}
for (int i = 0; i < value; i++) {
int number = 0;
number+= 2;
//print spaces v v v
for (int x = 0; x < (value - value + i + 1); x++) {
cout << " ";
}
//print * v v v
for (int y = 0; y < (/*ATTENTION: What do I do here? Plz help*/); y++) {
cout << "*";
}
cout << endl;
}
return 0;
}
What I've been trying to do is figure out what to put inside the parenthesis where it says (//ATTENTION). I've been working for at least an hour trying to do random things, and one time it worked when I input 4, but not for 5, and it's just been very hard. This is key to building the diamond, try putting in just value and compile to see what happens. I need it to be symmetrical.
I need to know what to put inside the parenthesis please. I'm sorry this is very long but the help would be appreciated thanks.
I also apologize if my code is messy and hard to read.
int number = 0; and number+= 2;
value - value inside for (int x = 0; x < (value - value + i + 1); x++) {
are not required.
Inside the parenthesis, you can use
2*(value-i-1)-1
However, I would suggest you to first analyze the problem and then try to solve it instead of trying random things. For instance, let's consider the cases of even and odd inputs i.e., 2 and 3.
Even Case (2)
*
***
***
*
The Analysis
Row Index Number of Spaces Number of Stars
0 1 1
1 0 3
2 0 3
3 1 1
For row index < value
Number of Spaces = value - row index - 1
Number of Stars = 2 * row index + 1
For row index >=value
The number of spaces and stars are simply reversed. In the odd cases, the situation is similar too with a small exception.
Odd Case (3)
*
***
*****
***
*
The Analysis
Row Index Number of Spaces Number of Stars
0 2 1
1 1 3
2 0 5
3 1 3
4 2 1
The small exception is that while reversing, we have to ignore the row index = value.
Now, if we put the above analysis in code we get the solution
//Define the Print Function
void PrintDiamond(int rowIndex, int value)
{
//print spaces v v v
for (int x = 0; x < value - rowIndex -1; x++) {
cout << " ";
}
//print * v v v
for (int y = 0; y < 2 * rowIndex + 1; y++) {
cout << "*";
}
cout << endl;
}
And then inside main
//Row index < value
for (int i = 0; i < value; i++) {
PrintDiamond(i,value);
}
//For row index >= value reversing the above case
//value-(value%2)-1 subtracts 1 for even and 2 for odd cases
//ignore the row index = value in odd cases
for (int i = value-(value%2)-1; i >=0; i--) {
PrintDiamond(i,value);
}
#include <iostream>
#include <sstream>
using namespace std;
int main() {
int value = 0;
cout << "Please enter in a value: ";
cin >> value;
cout << endl;
for (int i = 0; i < value; i++) {
//print spaces v v v
for (int x = 0; x < (value - i - 1); x++) {
cout << " ";
}
//print * v v v
for (int y = 0; y < (2 * i + 1); y++) {
cout << "*";
}
cout << endl;
}
for (int i = 0; i < value-1; i++) {
// int number = 0;
// number+= 2;
// //print spaces v v v
for (int x = 0; x < i+1; x++) {
cout << " ";
}
//print * v v v
for (int y = 0; y < (2*(value-1-i)-1); y++) {
cout << "*";
}
cout << endl;
}
return 0;
}
I hope that you will get this .Also in the second for loop you were iterating it one extra time by iterating the loop upto value. But since the pyramid is symmetric so the no of rows in the pyramid will be 2*value-1.So I in the second loop i should vary upto value -1.
This code should resolve the problem:
#include <sstream>
using namespace std;
void printSpaces(int howMany) {
for(int i = 0; i < howMany; i++) cout << " ";
}
void figure(int size) {
bool oddSize = size % 2 == 1;
int center = size / 2;
int spaces = size / 2;
// If figure is of an odd size adjust center
if (oddSize) {
center++;
} else { // Else if figure is of even size adjust spaces
spaces--;
}
for (int i = 1; i <= center; i++) {
printSpaces(spaces);
for(int j = 0; j < 1 + (i - 1) * 2; j++) cout << "*";
cout << endl;
spaces--;
}
spaces = oddSize ? 1 : 0; // If the figure's size is odd number adjust spaces to 1
center -= oddSize ? 1 : 0; // Adjust center if it's an odd size figure
for(int i = center; i >= 1; i--) {
printSpaces(spaces);
for(int j = 0; j < 1 + (i - 1) * 2; j++)
cout << "*";
cout << endl;
spaces++;
}
}
int main() {
int value = 0;
while(value < 3) {
cout << "Please enter in a value (>= 3): ";
cin >> value;
cout << endl;
}
figure(value);
return 0;
}
I'm working on a code that finds all saddle points in a matrix. Both smallest in their row and biggest in their column, and biggest in their row and smallest in their column fall under the definition (of my university) of a saddle point. Being a beginner I managed to get half of it done (finding saddle points which are smallest in their row and biggest in their column) by copying parts of what we've done in class and typing it myself. I have been stuck on it for quite some time and can't figure how to add the saddle points which are biggest in their row and smallest in their column to the program.
This is what I have so far:
#include <iostream>
#include <cstdlib>
using namespace std;
int a[10][10];
int x, y;
int pos_max(int j) //saddle points check
{
int max = 0;
for (int i = 1; i <= x - 1; i++) {
if (a[i][j] > a[max][j]) {
max = i;
}
}
return max;
}
int main() {
cout << "Enter the number of rows: ";
cin >> x;
cout << "Enter the number of columns: ";
cin >> y;
cout << "----------------------------" << endl;
for (int i = 0; i <= x - 1; i++) //input of the matrix
for (int j = 0; j <= y - 1; j++) {
cout << "a[" << i + 1 << ", " << j + 1 << "] = ";
cin >> a[i][j];
}
cout << "----------------------------\n";
for (int i = 0; i <= x - 1; i++) //visualization of the matrix
{
for (int j = 0; j <= y - 1; j++)
cout << a[i][j] << " ";
cout << endl;
}
cout << "----------------------------\n";
int r;
int flag = 0;
int i = y;
for (int j = 0; j <= y - 1; j++) {
r = pos_max(j);
for (i = 0; i <= y - 1; i++) {
if (a[r][i] < a[r][j]) {
break;
}
}
if (i == y) {
cout << "Saddle points are: ";
cout << "a[" << r + 1 << ", " << j + 1 << "] = " << a[r][j] << "\n";
flag = 1;
}
}
if (flag == 0) {
cout << "No saddle points\n";
}
cout << "----------------------------\n";
return 0;
}
First, there is a logical error with your code. In the pos_max function, it will return the index of the element which is maximum in the column. There can be a case when there are multiple maximum with the same value in the column, however, it returns the one which is not the minimum in the row, hence your program won't be able to print that saddle point.
To solve this, you can either return an array of all indices which are maximum in a column and then check for each of those points if it's minimum in their respective column, but I think it's not a very elegant solution. In any case, you will again have to write the entire code for the other condition for saddle points, minimum in column and maximum in row.
Hence, I would suggest a change in strategy. You create 4 arrays, max_row, max_col, min_row, min_col, where each array stores the minimum / maximum in that row / column respectively. Then you can traverse the array and check if that point satisfies saddle point condition.
Here is the code:
#include <iostream>
#include <cstdlib>
using namespace std;
int a[10][10];
int max_row[10], max_col[10], min_row[10], min_col[10];
int x, y;
bool is_saddle(int i, int j) {
int x = a[i][j];
return (max_row[i] == x && min_col[j] == x) || (min_row[i] == x && max_col[j] == x);
}
int main() {
/* code to input x, y and the matrix
...
*/
/* code to visualize the matrix
...
*/
/* populating max and min arrays */
for (int i = 0; i <= x-1; ++i) {
max_row[i] = a[i][0], min_row[i] = a[i][0];
for (int j = 0; j <= y-1; ++j) {
max_row[i] = max(max_row[i], a[i][j]);
min_row[i] = min(min_row[i], a[i][j]);
}
}
for (int j = 0; j <= y-1; ++j) {
max_col[j] = a[0][j], min_col[j] = a[0][j];
for (int i = 0; i <= x-1; ++i) {
max_col[j] = max(max_col[j], a[i][j]);
min_col[j] = min(min_col[j], a[i][j]);
}
}
/* Check for saddle point */
for (int i = 0; i <= x-1; ++i) {
for (int j = 0; j <= y-1; ++j) {
if (is_saddle(i, j)) {
cout << "Saddle points are: ";
cout << "a[" << i + 1 << ", " << j + 1 << "] = " << a[i][j] << "\n";
flag = 1;
}
}
}
if (flag == 0) {
cout << "No saddle points\n";
}
cout << "----------------------------\n";
return 0;
}
#include <iostream>
using namespace std;
int getMaxInRow(int[][5], int, int, int);
int getMinInColumn(int[][5], int, int, int);
void getSaddlePointCordinates(int [][5],int ,int );
void getInputOf2dArray(int a[][5], int, int);
int main()
{
int a[5][5] ;
int rows, columns;
cin >> rows >> columns;
getInputOf2dArray(a, 5, 5);
getSaddlePointCordinates(a,rows,columns);
}
void getInputOf2dArray(int a[][5], int rows, int columns)
{
for (int i = 0; i < rows; i = i + 1)
{
for (int j = 0; j < columns; j = j + 1)
{
cin >> a[i][j];
}
}
}
void getSaddlePointCordinates(int a[][5],int rows,int columns)
{
int flag = 0;
for (int rowNo = 0; rowNo < 5; rowNo++)
{
for (int columnNo = 0; columnNo < 5; columnNo++)
{
if (getMaxInRow(a, rows, columns, rowNo) == getMinInColumn(a, rows, columns, columnNo))
{
flag = 1;
cout << rowNo << columnNo;
}
}
}
if (flag == 0)
cout << "no saddle point";
cout << "\n";
}
int getMaxInRow(int a[][5], int row, int column, int rowNo)
{
int max = a[rowNo][0];
for (int i = 1; i < column; i = i + 1)
{
if (a[rowNo][i] > max)
max = a[rowNo][i];
}
return max;
}
int getMinInColumn(int a[][5], int row, int column, int columnNo)
{
int min = a[0][columnNo];
for (int i = 1; i < row; i = i + 1)
{
if (a[i][columnNo] < min)
min = a[i][columnNo];
}
return min;
}
just take the reference arr(ref[size]) // memorization method to check the minimum and maximum value in it.
Here is the Code Implementation with time complexity O(n *n) & space complexity O(n):
#include <bits/stdc++.h>
using namespace std;
#define size 5
void util(int arr[size][size], int *count)
{
int ref[size]; // array to hold all the max values of row's.
for(int r = 0; r < size; r++)
{
int max_row_val = arr[r][0];
for(int c = 1; c < size; c++)
{
if(max_row_val < arr[r][c])
max_row_val = arr[r][c];
}
ref[r] = max_row_val;
}
for(int c = 0; c < size; c++)
{
int min_col_val = arr[0][c];
for(int r = 1; r < size; r++) // min_val of the column
{
if(min_col_val > arr[r][c])
min_col_val = arr[r][c];
}
for(int r = 0; r < size; r++) // now search if the min_val of col and the ref[r] is same and the position is same, if both matches then print.
{
if(min_col_val == ref[r] && min_col_val == arr[r][c])
{
*count += 1;
if((*count) == 1)
cout << "The cordinate's are: \n";
cout << "(" << r << "," << c << ")" << endl;
}
}
}
}
// Driver function
int main()
{
int arr[size][size];
for(int i = 0; i < size; i++)
{
for(int j = 0; j < size; j++)
cin >> arr[i][j];
}
int count = 0;
util(arr, &count);
if(!count)
cout << "No saddle points" << endl;
}
// Test case -> Saddle Point
/*
Input1:
1 2 3 4 5
6 7 8 9 10
1 2 3 4 5
6 7 8 9 10
0 2 3 4 5
Output1:
The cordinate's are:
(0,4)
(2,4)
(4,4)
Input2:
1 2 3 4 5
6 7 8 9 1
10 11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
Output2:
No saddle points
*/
I want to convert a number into binary and store those binary value:
Example:
11 is 1011.. so I want to store those 2^3, 2^1, 2^1 numbers, so I can use those number. but my code doesn't store them right correctly. I got 8 4 2 for that instead of 8 2 1
Here is my code:
using namespace std;
int main() {
int num, arr[64];
int binary[10000];
cin >> num;
int expmax;
expmax = log2(num);
cout << expmax << endl;
int b = expmax;
int i = 0, r;
while (num != 0)
{
r = num % 2;
arr[i++] = r;
num /= 2;
}
for (int m = 0; m <= expmax ; m++)
{
cout << "array " << arr[m] << endl;
}
cout << endl;
int n = 0;
for (n++; n <= expmax; expmax--)
{
if (arr[n] = 1)
{
binary[n] = pow(2, expmax);
cout << binary[n] << endl;
}
else
{
binary[n] = pow(2, expmax)*0;
cout << binary[n] << endl;
}
}
cout << endl;
for (int j = i - 1; j >= 0; j--)
{
cout << arr[j];
}
cout << endl;
}
Your statement:
if (arr[n] = 1)
sets arr[n] to 1 and is always true (it evaluates to 1).
You want:
if (arr[n] == 1)
which is only true if arr[n] equals 1.
You are also bumping n to 1 at the start of the loop with:
int n = 0;
for (n++; n <= expmax; expmax--)
^^
Use this instead:
for (int n = 0; n <= expmax; ++n)
So, I have the following problem:
From the file tabl.in a number n will be read (n<=50).
After that a square array with n rows and n columns will be read; all the numbers in the array will be composed by a maximum of 2 digits each.
Shown in the file tabl.out, the modulo between the sum of numbers found on the second diagonal of the array and 10, if the sum is palindrome (true=1, false=0), and the arithmetic mean of elements situated below of the main diagonal.
Will be writing functions for:
reading the array
calculation of the operation sum of secondary diagonal%10
checking if the previous result it is palindrome
calculation of the arithmetic mean below main diagonal
Example:
tabl.in:
4
5 8 2 12
1 0 3 16
1 2 1 11
5 7 2 19
tabl.out:
2 1 3
where
(12+3+2+5)%10 = 22%10 = 2
22 is palindrome = 1
1+2+2+1+7+5 = 18, 18/6=3
My code so far is:
#include <fstream>
using namespace std;
ifstream fin("tabl.in");
ofstream fout("tabl.out");
void readn(int Arr[][51], int n) {
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
fin >> Arr[i][j];
}
int sumsec(int Arr[][51], int n) {
int s = 0;
float r;
for (int i = 1; i <= n; i++)
s = s + Arr[i][n - i + 1];
r = s % 10;
return r;
}
void pald(int Arr[][51], int n) {
int s = 0, pal = 0;
for (int i = 1; i < n; i++)
s = s + Arr[i][n - i + 1];
while (s != 0) {
pal = pal * 10 + s % 10;
s = s / 10;
}
if (pal == s)
fout << "1 ";
else
fout << "0 ";
}
int ambmd(int Arr[][51], int n) {
int s = 0, k;
float ame;
for (int i = 2; i <= n; i++) {
for (int j = 1; j <= i - 1; j++) {
s = s + Arr[i][j];
k++;
}
}
ame = s / k;
return ame;
}
int main() {
int Arr[51][51], n;
float r, ame;
fin >> n;
readn(Arr, n);
r = sumsec(Arr, n);
fout << r << " ";
pald(Arr, n);
ame = ambmd(Arr, n);
fout << ame;
}
But I have an issue with the palindrome() function: my output file will have 2 0 3 written to it for the given array from the example, instead of 2 1 3. What am I doing wrong?
Your pald function would work, if you compute s the same way as you do in sumsec and if s would still contain the sum, after you compute pal.
In your case, while (s != 0) {...}, followed by if (pal == s) {...} could be re-written as if (pal == 0), which is clearly not the intended solution. Just save your sum before computing pal, then compare with the saved sum.
Also, change your loop condition for computing s to for (int i = 1; i <= n; i++).
int s = 0, pal = 0, sum = 0;
for (int i = 1; i <= n; i++)
s = s + Arr[i][n - i + 1];
sum = s;
while (s != 0) {
pal = pal * 10 + s % 10;
s = s / 10;
}
if (pal == sum)
fout << "1 ";
else
fout << "0 ";
You should also consider the various comments for code improvements, like not re-computing the sum in the pald function.