Pascal's Triangle using mainly functions in C++ - c++

Question

I was trying to write a code that would display pascals triangle. Instead of displaying the result as :
my result is displayed as
1
1 1
1 2 1
1 3 3 1
Please help me figure out how to modify it to be able to get the actual triangle. I cant use arrays and pointers since those aren't covered in my class yet. Here's my code:
#include "stdafx.h"
#include <iostream>
using namespace std;
void PascalsTriangle(int);
int main()
{
int n;
cout << "Enter the number of rows you would like to print for Pascal's Triangle: ";
cin >> n;
PascalsTriangle(n);
return 0;
}
void PascalsTriangle (int n){
int i,j,x;
for(i=0;i<n;i++)
{
x=1;
for(j=0;j<=i;j++)
{
cout << x << " ";
x = x * (i - j) / (j + 1);
}
cout << endl;
}
}

Answer 1

Here is the updated version. Works for any n. I've added a function to return the number of digits in a number since that computation is needed each iteration of the inner loop.
#include <iostream>
#include <string>
using namespace std;
void PascalsTriangle(int);
int main()
{
int n;
cout << "Enter the number of rows you would like to print for Pascal's Triangle: ";
cin >> n;
cout << endl;
PascalsTriangle(n);
return 0;
}
int numdigits(int x)
{
int count = 0;
while(x != 0) {
x = x / 10;
++count;
}
return count;
}
void PascalsTriangle (int n)
{
int i, j, x, y, maxlen;
string len;
for(i = 0; i < n; i++) {
x = 1;
len = string((n-i-1)*(n/2), ' ');
cout << len;
for(j = 0; j <= i; j++) {
y = x;
x = x * (i - j) / (j + 1);
maxlen = numdigits(x) - 1;
if(n % 2 == 0)
cout << y << string(n - 1 - maxlen, ' ');
else {
cout << y << string(n - 2 - maxlen, ' ');
}
}
cout << endl;
}
}
OUTPUTS:
Enter the number of rows you would like to print for Pascal's Triangle: 3
1
1 1
1 2 1
Enter the number of rows you would like to print for Pascal's Triangle: 6
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Enter the number of rows you would like to print for Pascal's Triangle: 9
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
Enter the number of rows you would like to print for Pascal's Triangle: 12
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
UPDATE for a more tighter triangle:
void PascalsTriangle(int n)
{
int i, j, x, y, maxlen;
string len;
for(i = 0; i < n; i++) {
x = 1;
if(n % 2 != 0)
len = string((n-i-1)*(n/2), ' ');
else
len = string((n-i-1)*((n/2)-1), ' ');
cout << len;
for(j = 0; j <= i; j++) {
y = x;
x = x * (i - j) / (j + 1);
maxlen = numdigits(x);
if(n % 2 == 0)
cout << y << string(n - 2 - maxlen, ' ');
else {
cout << y << string(n - 1 - maxlen, ' ');
}
}
cout << endl;
}
}
OUTPUT
Enter the number of rows you would like to print for Pascal's Triangle: 3
1
1 1
1 2 1
Enter the number of rows you would like to print for Pascal's Triangle: 6
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Enter the number of rows you would like to print for Pascal's Triangle: 9
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
Enter the number of rows you would like to print for Pascal's Triangle: 12
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1

Answer 2

Try this :
#include <iostream>
using namespace std;
int main()
{
int n,k,i,x;
cout << "Enter a row number for Pascal's Triangle: ";
cin >> n;
for(i=0;i<=n;i++)
{
x=1;
for(k=0;k<=i;k++)
{
cout << x << " ";
x = x * (i - k) / (k + 1);
}
cout << endl;
}
return 0;
}
EDITED:
#include <iostream>
using namespace std;
int main()
{
int n,coef=1,space,i,j;
cout<<"Enter number of rows: ";
cin>>n;
for(i=0;i<n;i++)
{
for(space=1;space<=n-i;space++)
cout<<" ";
for(j=0;j<=i;j++)
{
if (j==0||i==0)
coef=1;
else
coef=coef*(i-j+1)/j;
cout<<" "<<coef;
}
cout<<endl;
}
}

Answer 3

Do you expect this result?
1111
123
13
1
You need to add endl or constant string equivalent to "\r\n" to the end of the output string, or use commas in output, then, for example:
"1111,123,13,1"
"1,11,121,1331"

Answer 4

If you want it to look like a 'triangle', that is, a symmetric looking isosceles triangle, try this code for your PascalTriangle function. The only problem with this is that when you get larger digits, it will break some of the symmetry but up to 5 rows it'll work fine.
void PascalsTriangle(int n)
{
int i, j, x;
string len;
for(i = 0; i < n; i++)
{
x = 1;
len = string(n - i - 1, ' ');
cout << len;
for(j = 0; j <= i; j++)
{
cout << x << " ";
x = x * (i - j) / (j + 1);
}
cout << endl;
}
}
OUTPUT
Enter the number of rows you would like to print for Pascal's Triangle: 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
As opposed to:
Enter the number of rows you would like to print for Pascal's Triangle: 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

Answer 5

#include <iostream>
using namespace std;
const int MAXC = 13; //-- Max column constant
const int MAXF = 7; //-- Max row constant
int m, n, k, x, y; //-- Counters and accumulators for loops
int arrayTriangulo[MAXF][MAXC]; // Array that stores the values of the Pascal Triangle
int main() {
m = (MAXC/2); // Middle Column
k = (MAXC/2); // Middle row
//-- 1.- Fill in the Array from Left to Right and from Top to Bottom
//-- 2.- Fill in the Array starting from row 0 through 13 (MAXF)
for ( x = 0; x < MAXF; x++ ) {
n = 1;
//-- 3.- Fill in the Array starting from Column 0 through 7 (MAXC)
for ( y = 0; y < MAXC; y++ ) {
//-- Assign 0 to the Array element that is not part of the triangle.
arrayTriangulo[x][y] = 0;
//-- 4.- If it is on the edges of the triangle assigns the value of "n". Which we initialize in 1.
if (( y == m ) || ( y == k )) {
arrayTriangulo[x][y] = n;
}
//-- 5.- For the rest of the internal values of the triangle other than the edges.
//-- The sum of the value is assigned (upper left row -1) + (upper right row + 1)
if ( ( x > 1 ) && ( x < MAXF ) && ( y < MAXC-1 ) ) {
arrayTriangulo[x][y] = arrayTriangulo[x-1][y-1] + arrayTriangulo[x-1][y+1];
}
//-- 6.- Finally Draw the Triangle by omitting the values at zero.
if ( arrayTriangulo[x][y] > 0 )
cout << arrayTriangulo[x][y] << " ";
else
cout << " ";
}
cout << endl;
m--;
k++;
}
return 0;
}