Assignment:
Write a program that prompts the user for an integer value representing the height of a triangle. The program should then print out a triangle of O’s of that height, with a vertically aligned right edge.
My problem:
I have figured out the code to execute a normal triangle, but I am having some difficulty on writing code to have spacing before my "0's" to make it aligned to the right.
#include <iostream>
using namespace std;
int main()
{
int triHeight;
int c = 0;
int r = 0;
int k = 0;
cout << "Enter the triangle height: " << endl;
cin >> triHeight;
for (c = 0; c <= triHeight; c = c+1)
{
for (r = 0; r < c; r = r + 1)
{
cout << "0";
}
for (k = 0; k <= c; k = k - 1)
{
cout << " ";
}
cout << endl;
}
system("pause");
return 0;
}
You have two main problems :
you write the spaces after the 0 rather than before to have the 0 indented
for (k = 0; k <= c; k = k - 1) never ends up to the possible effect of an overflow
I also encourage you to check the result of >> to be sure a valid integer was enter so like if (!(cin >> triHeight)) cerr << "invalid height" << endl; else { ... }
A right way close to yours using loops for all is (supposing you want a pyramid) :
#include <iostream>
using namespace std;
int main()
{
int triHeight;
cout << "Enter the triangle height: " << endl;
if (!(cin >> triHeight))
cerr << "invalid height" << endl;
else {
for (int h = 1; h <= triHeight; h += 1) {
for (int s = triHeight - h; s != 0; s -= 1)
cout << ' ';
for (int z = 2*(h-1)+1; z >0; z -=1)
cout << '0';
cout << endl;
}
}
}
Compilation and execution :
/tmp % g++ -pedantic -Wextra -Wall c.cc
/tmp % ./a.out
Enter the triangle height:
5
0
000
00000
0000000
000000000
/tmp %
If you want half a pyramid :
#include <iostream>
using namespace std;
int main()
{
int triHeight;
cout << "Enter the triangle height: " << endl;
if (!(cin >> triHeight))
cerr << "invalid height" << endl;
else {
for (int h = 1; h <= triHeight; h += 1) {
for (int s = triHeight - h; s != 0; s -= 1)
cout << ' ';
for (int z = 0; z < h; z += 1)
cout << '0';
cout << endl;
}
}
}
Compilation and execution :
/tmp % g++ -pedantic -Wall -Wextra c.cc
/tmp % ./a.out
Enter the triangle height:
5
0
00
000
0000
00000
/tmp %
You can also do not make the two internal loops by yourself :
#include <iostream>
using namespace std;
int main()
{
int triHeight;
cout << "Enter the triangle height: " << endl;
if (!(cin >> triHeight))
cerr << "invalid height" << endl;
else {
for (int h = 1; h <= triHeight; h += 1)
cout << string(triHeight - h, ' ') << string(h, '0') << endl;
}
}
Compilation and execution :
/tmp % g++ -pedantic -Wall -Wextra c.cc
/tmp % ./a.out
Enter the triangle height:
5
0
00
000
0000
00000
/tmp %
that solution is shorter but creates temporary strings
The logic of your program is wrong. At first you are trying to output the symbol 'O'
for (r = 0; r < c; r = r + 1)
{
cout << "0";
}
and after that you are trying to output spaces
for (k = 0; k <= c; k = k - 1)
{
cout << " ";
}
Moreover this loop is invalid because the variable k is decremented starting from 0.
The program can be written using only one loop and standard i/o manipulators declared in the header <iomanip>
Here is a demonstrative program
#include <iostream>
#include <iomanip>
int main()
{
while ( true )
{
const char c = 'O';
std::cout << "Enter the height of a triangle (0 - exit): ";
int height = 0;
if ( not ( std::cin >> height ) or ( height <= 0 ) ) break;
std::cout << '\n';
for ( int i = 0; i < height; i++ )
{
std::cout << std::setw( height - i ) << std::setfill( ' ' ) << c;
std::cout << std::setw( i + 1 ) << std::setfill( c ) << '\n';
}
std::cout << '\n';
}
return 0;
}
The program output might look the following way
Enter the height of a triangle (0 - exit): 1
O
Enter the height of a triangle (0 - exit): 2
O
OO
Enter the height of a triangle (0 - exit): 3
O
OO
OOO
Enter the height of a triangle (0 - exit): 4
O
OO
OOO
OOOO
Enter the height of a triangle (0 - exit): 5
O
OO
OOO
OOOO
OOOOO
Enter the height of a triangle (0 - exit): 6
O
OO
OOO
OOOO
OOOOO
OOOOOO
Enter the height of a triangle (0 - exit): 7
O
OO
OOO
OOOO
OOOOO
OOOOOO
OOOOOOO
Enter the height of a triangle (0 - exit): 8
O
OO
OOO
OOOO
OOOOO
OOOOOO
OOOOOOO
OOOOOOOO
Enter the height of a triangle (0 - exit): 9
O
OO
OOO
OOOO
OOOOO
OOOOOO
OOOOOOO
OOOOOOOO
OOOOOOOOO
Enter the height of a triangle (0 - exit): 10
O
OO
OOO
OOOO
OOOOO
OOOOOO
OOOOOOO
OOOOOOOO
OOOOOOOOO
OOOOOOOOOO
Enter the height of a triangle (0 - exit): 0
The simplest way to do this is to use the built in features of the std::iostream:
#include <iostream>
#include <iomanip>
#include <string>
int main()
{
int triHeight;
int c = 0;
std::cout << "Enter the triangle height: \n";
std::cin >> triHeight;
std::cout << std::setfill(' ') << std::right;
for (c = 0; c < triHeight; c++)
{
std::cout << std::setw(triHeight) << std::string(c+1,'0') << '\n';
}
return 0;
}
You can get back to a left aligned triangle by simply changing std::right to std::left.
Related
When I compile my program, run it and enter values I will get a rather strange and unexpected output:
So if I enter:
Enter width and height: 9 5
Enter characters: X O
1 XOXOXOXOX
2 OXOXOXOXO
3 XOXOXOXOX
4 OXOXOXOXO
5 XOXOXOXOX
A BCDEFGHI
When it's supposed to be:
Enter width and height: 9 5
Enter characters: X O
1 XOXOXOXOX
2 OXOXOXOXO
3 XOXOXOXOX
4 OXOXOXOXO
5 XOXOXOXOX
ABCDEFGHI
When do my void print_alphabet in another program it will work out just fine so I don't know the problem. I believe it has something to do with my other function but I can not seem to get it to work. Why does it act that way? Why does it print out A and then it does setw and prints out the rest?
This is my code:
#include <iostream>
#include <iomanip>
using namespace std;
void print_chess_board (int const height,
int const width,
char const char_1,
char const char_2)
{
int index {};
for (int i = 1; i <= height; ++i)
{
if (i%2)
{
index = 0;
}
else
{
index = 1;
}
cout << left << setw(3) << i;
for (int j {}; j < width; ++j)
{
if (++index%2 == 0)
{
cout << char_2;
}
else
{
cout << char_1;
}
}
cout << endl;
}
}
void print_alphabet (int const width)
{
cout << setfill(' ') << setw(4);
for (int i {}; i < width; ++i)
{
cout << char('A' + i);
}
}
int main()
{
int width {};
int height {};
char char_1 {};
char char_2 {};
cout << "Enter width and height: ";
cin >> width >> height;
cout << "Enter characters: ";
cin >> char_1 >> char_2;
print_chess_board(height,width,char_1,char_2);
print_alphabet(width);
return 0;
}
You need to change
cout << setfill(' ') << setw(4);
to
cout << right << setfill(' ') << setw(4);
I am trying to build a triangle, with a user entered base and height.
When these entered values are different (base!=height), the program goes haywire and gets stuck in the triangle draw loop.
I've tried altering the code a couple of times, but please treat me as a programming novice.
//BUILD TRIANGLE//
#include <string>
#include <iomanip>
#include <iostream>
int main()
{
std::cout << "\nEnter base and height:\n";
int height{0}; int base{0};
std::cin >> base >> height;
std::string bottom(base, '*');
std::string top = "*";
int middlerows = height - 1;
int middlespacechars;
std::cout << top << std::endl;
for (middlespacechars = 0;
middlerows != 1 || middlespacechars != base - 2;
++middlespacechars, --middlerows) {
std::string middlespace(middlespacechars, ' ');
std::cout << "*" << middlespace << "*\n";
}
std::cout << bottom << "\n" << std::endl;
std::cout << "^TRIANGLE\n";
std::cout << "BASE = " << base << std::endl;
std::cout << "HEIGHT = " << height << std::endl;
std::cout << "goodbye" << "\n" << std::endl;
}
The output is totally haywire, with asterisks across the screen in no discernible shape.
When I put in values where base=height, though, a pretty little right angle triangle pops up.
With your code, you can only draw well triangles which have base equal to height.
If you change stop condition in your for loop, you can get what you probably want to get:
for (middlespacechars = 0; middlerows != 1 || middlespacechars != base - 2; ++middlespacechars, --middlerows) {
... into ...
for (middlespacechars = 0; middlerows > 1 || middlespacechars < base - 2; ++middlespacechars, --middlerows) {
It was huge probability that if base and height are different then stop condition will not be achieved. For loop in your code will stop if middlerows will be 1 and middlespacechars will be base - 2 at the same moment.
Test it here.
//C++ program to display hollow star pyramid
#include<iostream>
using namespace std;
int main()
{
int rows, i, j, space;
cout << "Enter number of rows: ";
cin >> rows;
for(i = 1; i <= rows; i++)
{
//for loop to put space in pyramid
for (space = i; space < rows; space++)
cout << " ";
//for loop to print star
for(j = 1; j <= (2 * rows - 1); j++)
{
if(i == rows || j == 1 || j == 2*i - 1)
cout << "*";
else
cout << " ";
}
cout << "\n";
}
return 0;
}
I need to Write a C ++ program that prompts the user to enter 10 integers in the array.
Input numbers are: 0, 1 or 2.
The program should extract the array that is entered on the screen Arrange the array of elements 0, 1, and 2 so that the array places all 0 in the first places, then all 1s
and all 2 as the last.
Arranged array displays on the screen.
Have been struggling over this for few hours now. Im stuck and dont know how to show Output .
Input should be for example 0 0 1 0 1 2 2 2 0 1
Output 0000111222
HOW?
int main ()
{
int t [N], i, ;
cout << "Enter 10 arrays from 0-2" << endl;
cout << "Original array:";
for (i = 0; i <N; i ++)
{
cin >> t [i];
}
if (t [i]> = 0 && t [i] <= 2)
{
cout << "Rearranging elements of array:" << ? << endl;
}
cout << "End of the program!"
return 0;
}
when you do
if (t [i]> = 0 && t [i] <= 2)
i equals N so you access out of the array, and of course that does not sort the array
You do not check if cin >> t[i] success so if the user enter something else than an int all the entries from the current will not be set (they will be 0 in case you use a std::vector)
A first way is to do without taking account the range 0..2, replace int t[n] by std::vector<int> t(N) and use sort(t.begin(), t.end()) to sort your array
The complexity is O(N*log(N)) (here N is 10)
For instance :
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 10
int main ()
{
vector<int> t(N);
size_t i; // size_t the right type for an index
cout << "Enter 10 values in range 0..2" << endl;
for (i = 0; i < N; ++i)
{
for (;;) {
cout << "value #" << i << ':';
if (!(cin >> t[i])) {
cerr << "not a number" << endl;
cin.clear(); // raz error
string s;
cin >> s; // skip bad input
}
else if ((t[i] < 0) || (t[i] > 2))
cerr << "value out of range" << endl;
else
break;
}
}
cout << "Original array:";
for (i = 0; i < N; ++i) cout << ' ' << t[i]; // old way to do
cout << endl;
sort(t.begin(), t.end());
cout << "Sorted array:";
for (auto v : t) cout << ' ' << v; // new way to do
cout << endl;
cout << "End of the program!" << endl;
return 0;
}
Compilation and execution :
pi#raspberrypi:/tmp $ g++ -pedantic -Wextra -Wall s.cc
pi#raspberrypi:/tmp $ ./a.out
Enter 10 values in range 0..2
value #0:aze
not a number
value #0:-2
value out of range
value #0:3
value out of range
value #0:2
value #1:0
value #2:1
value #3:2
value #4:0
value #5:2
value #6:1
value #7:0
value #8:0
value #9:1
Original array: 2 0 1 2 0 2 1 0 0 1
Sorted array: 0 0 0 0 1 1 1 2 2 2
End of the program!
A second way considering a range [min .. max] not too large is to count the number of each value then fill the array to respect these counts
The complexity is O(2N) (here N is 10)
For instance :
#include <vector>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
#define MIN 0
#define MAX 2
#define N 10
int main ()
{
vector<int> t(N);
size_t i; // size_t the right type for an index
cout << "Enter 10 values in range " << MIN << ".." << MAX << endl;
for (i = 0; i < N; ++i)
{
for (;;) {
cout << "value #" << i << ':';
if (!(cin >> t[i])) {
cerr << "not a number" << endl;
cin.clear(); // raz error
string s;
cin >> s; // skip bad input
}
else if ((t[i] < MIN) || (t[i] > MAX))
cerr << "value out of range" << endl;
else
break;
}
}
cout << "Original array:";
for (auto v : t) cout << ' ' << v;
cout << endl;
// count numbers
vector<size_t> counts(MAX - MIN + 1);
for (auto v : t) counts[v - MIN] += 1;
// fill again
i = 0;
for (int r = MIN; r <= MAX; ++r) {
size_t n = counts[r - MIN];
while (n--) t[i++] = r;
}
cout << "Sorted array:";
for (auto v : t) cout << ' ' << v;
cout << endl;
cout << "End of the program!" << endl;
return 0;
}
Compilation and execution :
pi#raspberrypi:/tmp $ g++ -pedantic -Wextra -Wall s2.cc
pi#raspberrypi:/tmp $ ./a.out
Enter 10 values in range 0..2
value #0:a
not a number
value #0:3
value out of range
value #0:0
value #1:2
value #2:1
value #3:1
value #4:2
value #5:2
value #6:2
value #7:0
value #8:1
value #9:2
Original array: 0 2 1 1 2 2 2 0 1 2
Sorted array: 0 0 1 1 1 2 2 2 2 2
End of the program!
Specifically for values between 0 and 2 (in fact for 3 possible values) as said in a remark by #PaulMcKenzie you can use the Dutch national flag problem and look at that question : R G B element array swap
The complexity is O(N) (here N is 10)
As a beginner, you may want to try this (this is the simplest solution I could think of without using other libraries):
int main () {
const int size = 10;
int arr[size], temp = 0;
for (int i = 0; i < size; i++) {
cin >> arr[i];
}
for (int j = 0; j < size - 1; j++) {
for (int k = 0; k < size - j - 1; k++) {
if(arr[k] > arr[k+1]) {
temp = arr[k];
arr[k] = arr[k+1];
arr[k+1] = temp;
}
else
continue;
}
}
for (int i = 0; i < size; i++) {
cout << arr[i] << " ";
}
return 0;
}
I hope this helps you.
1st answer is too difficult for me, im only beginner not knowing what im doing yet.
2nd answer is to that direction where it should be right but i need to put in that input cant be more than 2 or less than numeber 0, thats the problem now :D
Im sorry, i just cant get to that point where i understand that syntax.
I have to input a value in the program and keep dividing it by 4 until it reaches the number 0. But when I run it, it doesn't stop at 0, it keeps repeating 0 forever. What is wrong with the code?
#include <iostream>
using namespace std;
int main(){
double input;
cout << "Enter an Integer: ";
cin >> input;
cout << input << "/ 4 ";
do
{
input = input / 4;
if (input >= 0)
cout <<" = "<< input << endl;
cout <<input << " /4";
}
while ((input >= 0) || (input != 0));
return 0;
}
Here are my three cents.:)
#include <iostream>
int main()
{
const long long int DIVISOR = 4;
while ( true )
{
std::cout << "Enter an Integer (0 - Exit): ";
long long int n;
if ( not ( std::cin >> n ) or ( n == 0 ) ) break;
std::cout << std::endl;
do
{
std::cout << n << " / " << DIVISOR;
n /= DIVISOR;
std::cout << " = " << n << std::endl;
} while ( n );
std::cout << std::endl;
}
return 0;
}
The program output might look like
Enter an Integer (0 - Exit): 1000
1000 / 4 = 250
250 / 4 = 62
62 / 4 = 15
15 / 4 = 3
3 / 4 = 0
Enter an Integer (0 - Exit): 0
I am trying to make this shape from the code below. I'm confused as to how to make it print the 2nd row, second to last star without it skipping and printing the extra space before printing the star. Once that is figured out would the bottom half, when the stars expands back out, would the code be similar to the top half? I have tried a couple combinations of code between c and r but I have been stuck with what I currently.
---------------------- //row 0
* *| //row 1
* * * *| //row 2
* * * * * *|
* * * * * * * *|
* * * * * * * * * *|
* * * * * * * * * * *|
* * * * * * * * * *|
* * * * * * * *|
* * * * * *|
* * * *|
* *|
----------------------
#include <iostream>
using std::cout; using std::cin; using std::endl;
int main() {
cout << "Enter a positive odd number less than 40: ";
int num = 0;
int z = 1;
for (int a = 0; a < 3; ++a)
{
cin >> num;
if (num < 38 && num > 0 && num % 2 == 1)
{
cout << "Thank you!" << endl << endl;
for (int r = 0; r < num; ++r) //outer loop/rows
{
for (int c = 0; c < num; ++c) //inner loop/columns
{
if (r == 0) cout << "--"; //top of square
else if (c >= r + r - c && c < num - 1)
cout << " ";
//else if (c == num - 1) cout << "*|";
else if (r == num - 1) cout << "--"; //bottom of square
else if (c == num - 1) cout << "*|"; //right side of square
else if (r > c) cout << "* ";
}
cout << endl;
}
break;
}
else cout << "Please enter a positve odd number that is less than 40!" << endl;
}
cout << endl;
}
I just took two variables left=0 & right=num-1 and increased left & decreased right till r<=num/2, after that i reversed the process,when the col <= left or col >=right I printed *.
I hope it will be easy to understand.
Here is the code:
#include <iostream>
using std::cout; using std::cin; using std::endl;
int main() {
cout << "Enter a positive odd number less than 40: ";
int num = 0;
int z = 1;
for (int a = 0; a < 3; ++a)
{
cin >> num;
if (num < 38 && num > 0 && num % 2 == 1)
{
cout << "Thank you!" << endl << endl;
int left=0,right=num-1;
//for printing top line
for(int i = 0; i < num; i++) cout<<"- ";
cout<<"-"<<endl;
for (int r = 0; r < num; ++r) //outer loop/rows
{
//printing columns
for(int c = 0; c < num; c++)
{
if(c <= left || c >= right)
cout<<"* ";
else
cout<<" ";
}
if(r >= num/2) //checking for half of the rows
{
left--;right++;
}
else
{
left++;right--;
}
cout<<"|"<<endl;
}
//for printing last additional line
for(int i = 0; i < num; i++) cout<<"- ";
cout<<"-"<<endl;
break;
}
else cout << "Please enter a positve odd number that is less than 40!" << endl;
}
cout << endl;
}
This approach does it the math way.
Furthermore it draws a full frame with plus-chars at the edges.
Give it a try.
#include <iostream>
#include <cmath>
using std::cout; using std::cin; using std::endl;
int main() {
cout << "Enter a positive odd number less than 40: ";
int num = 0;
int z = 1;
for (int a = 0; a < 3; ++a) {
cin >> num;
if (num < 40 && num > 0 && num % 2 == 1) {
cout << "Thank you!" << endl << endl;
int center = ceil(num / 2.0);
for (int r = 0; r <= num+1; ++r) { //outer loop/rows
for (int c = 0; c <= num+1; ++c) { //inner loop/columns
if (r == 0 || r == num+1) {
if (c == 0 || c == num+1)
cout << "+"; // corner
else
//top or botton of square between corners
if (c == center)
cout << "-";
else
cout << "--";
}
else if (c == 0 || c == num+1) {
cout << "|"; // left or right frame
} else {
// inner part
if ((center-std::abs(center-r)) >= center-std::abs(center-c))
if (c < center)
cout << "* ";
else if (c > center)
cout << " *";
else
cout << "*";
else
if (c == center)
cout << " ";
else
cout << " ";
}
}
cout << endl;
}
} else
cout << "Please enter a positve odd number that is less than 40!" << endl;
}
cout << endl;
}
Just another way (with some more user input checking):
#include <iostream>
#include <string>
#include <limits>
#include <sstream>
using std::cout;
using std::cin;
using std::string;
const auto ssmax = std::numeric_limits<std::streamsize>::max();
const int max_dim = 40;
const int max_iter = 3;
int main() {
cout << "Enter a positive odd number less than " << max_dim << ": ";
int num = 0, counter = 0;
while ( counter < max_iter ) {
cin >> num;
if ( cin.eof() )
break;
if ( cin.fail() ) {
cout << "Please, enter a number!\n";
cin.clear();
cin.ignore(ssmax,'\n');
}
if ( num < max_dim && num > 0 && num % 2 ) {
cout << "Thank you!\n\n";
//top line
string line(num * 2, '-');
cout << line << '\n';
for ( int r = 0, border = num - 1; r < num; ++r ) {
cout << '*';
for ( int c = 1; c < num; ++c ) {
if ( (c > r && c < border) || (c < r && c > border) )
cout << " ";
else
cout << " *";
}
// right border
cout << "|" << '\n';
--border;
}
//bottom line
cout << line << '\n';
++counter;
} else {
cout << "Please, enter a positive odd number that is less than 40!\n";
}
}
cout << std::endl;
}
Or my favorite:
// top line
string line = string(num * 2, '-') + '\n';
cout << line;
// inside lines
int r = 0, border = ( num - 1 ) * 2;
string inside = string(border + 1, ' ') + "|\n";
// top
while ( r < border ) {
inside[r] = '*';
inside[border] = '*';
r += 2;
border -= 2;
cout << inside;
}
// center line
inside[r] = '*';
cout << inside;
// bottom
while ( border > 0 ) {
inside[r] = ' ';
inside[border] = ' ';
r += 2;
border -= 2;
cout << inside;
}
//bottom line
cout << line;