We were asked to make a triangular no. pattern in c++ with min. loops. The triangle pattern is as follows:

____1_____
___2__3____
__4__5__6__
7__8__9__10

Code:

#include <iostream>

using namespace std;
int main() {
    int n=0, r=0, i=0;
    cout << "No. of rows: ";
    cin >> r;

    for( n=1; n<=r; n++) {
        for( i=1; i<=r-n; i++) {
            cout << "  ";
        }
        for( i=(n*(n-1)/2)+1; i<=(n*(n+1)/2); i++ ) {
            if( i<10 )
                cout << " " << i << "  ";
            else
                cout << i << "  ";
        }
        cout << "\n";
    }
    return 0;
}

Output:

Here's what I'd like reviewed:

1. Is it wise to use pattern generating formulas? for eg, for putting value of i in last loop, i used the formula for the pattern \$1,2,4,7\$... as \$\frac{n*(n-1)}{2}+1\$. Is it efficient? This involves four operations in each iteration, and it's fairly slow.
2. Is it possible to reduce the number of loops? Is it better to reduce variables or reduce loops?

We were asked to make a triangular number pattern in C++ with minimum loops:

____1_____
___2__3____
__4__5__6__
7__8__9__10

Code:

#include <iostream>

using namespace std;
int main() {
    int n=0, r=0, i=0;
    cout << "No. of rows: ";
    cin >> r;

    for( n=1; n<=r; n++) {
        for( i=1; i<=r-n; i++) {
            cout << "  ";
        }
        for( i=(n*(n-1)/2)+1; i<=(n*(n+1)/2); i++ ) {
            if( i<10 )
                cout << " " << i << "  ";
            else
                cout << i << "  ";
        }
        cout << "\n";
    }
    return 0;
}

Output:

Here's what I'd like reviewed:

1. Is it wise to use pattern-generating formulas? For example, for putting value of i in the last loop, I used the formula for the pattern \$1,2,4,7\$... as \$\frac{n*(n-1)}{2}+1\$. Is it efficient? This involves four operations in each iteration, and it's fairly slow.
2. Is it possible to reduce the number of loops? Is it better to reduce variables or reduce loops?

We were asked to make a triangular no. pattern in c++ with min. loops. The triangle pattern is as follows:

____1_____
___2__3____
__4__5__6__
7__8__9__10

My code:

#include <iostream>

using namespace std;
int main() {
    int n=0, r=0, i=0;
    cout << "No. of rows: ";
    cin >> r;

    for( n=1; n<=r; n++) {
        for( i=1; i<=r-n; i++) {
            cout << "  ";
        }
        for( i=(n*(n-1)/2)+1; i<=(n*(n+1)/2); i++ ) {
            if( i<10 )
                cout << " " << i << "  ";
            else
                cout << i << "  ";
        }
        cout << "\n";
    }
    return 0;
}

OUTPUT

QUESTIONS

1) Is it wise to use pattern generating formulas? for eg, for putting value of i in last loop, i used the formula for the pattern \$1,2,4,7\$... as \$\frac{n*(n-1)}{2}+1\$. Is it efficient? I am asking this because this involves 4 operations in every iteration, which would make it prettu slow!

2) Is it possible to reduce no. of loops? is it better to reduce variables or reduce loops?

3) Can anyone show a recursive approach (or iterative)?

THANK YOU!

We were asked to make a triangular no. pattern in c++ with min. loops. The triangle pattern is as follows:

____1_____
___2__3____
__4__5__6__
7__8__9__10

Code:

#include <iostream>

using namespace std;
int main() {
    int n=0, r=0, i=0;
    cout << "No. of rows: ";
    cin >> r;

    for( n=1; n<=r; n++) {
        for( i=1; i<=r-n; i++) {
            cout << "  ";
        }
        for( i=(n*(n-1)/2)+1; i<=(n*(n+1)/2); i++ ) {
            if( i<10 )
                cout << " " << i << "  ";
            else
                cout << i << "  ";
        }
        cout << "\n";
    }
    return 0;
}

Output:

Here's what I'd like reviewed:

1. Is it wise to use pattern generating formulas? for eg, for putting value of i in last loop, i used the formula for the pattern \$1,2,4,7\$... as \$\frac{n*(n-1)}{2}+1\$. Is it efficient? This involves four operations in each iteration, and it's fairly slow.
2. Is it possible to reduce the number of loops? Is it better to reduce variables or reduce loops?

We were asked to make a triangular no. pattern in c++ with min. loops. The triangle pattern is as follows:

____1_____
___2__3____
__4__5__6__
7__8__9__10

My code:

#include <iostream>

using namespace std;
int main() {
    int n=0, r=0, i=0;
    cout << "No. of rows: ";
    cin >> r;

    for( n=1; n<=r; n++) {
        for( i=1; i<=r-n; i++) {
            cout << "  ";
        }
        for( i=(n*(n-1)/2)+1; i<=(n*(n+1)/2); i++ ) {
            if( i<10 )
                cout << " " << i << "  ";
            else
                cout << i << "  ";
        }
        cout << "\n";
    }
    return 0;
}

OUTPUT

QUESTIONS

1) Is it wise to use pattern generating formulas? for eg, for putting value of i in last loop, i used the formula for the pattern \$1,2,4,7\$... as \$\frac{n*(n-1)}{2}+1\$. Is it efficient?

2) Is it possible to reduce no. of loops? is it better to reduce variables or reduce loops?

3) Can anyone show a recursive approach (or iterative)?

THANK YOU!

We were asked to make a triangular no. pattern in c++ with min. loops. The triangle pattern is as follows:

____1_____
___2__3____
__4__5__6__
7__8__9__10

My code:

#include <iostream>

using namespace std;
int main() {
    int n=0, r=0, i=0;
    cout << "No. of rows: ";
    cin >> r;

    for( n=1; n<=r; n++) {
        for( i=1; i<=r-n; i++) {
            cout << "  ";
        }
        for( i=(n*(n-1)/2)+1; i<=(n*(n+1)/2); i++ ) {
            if( i<10 )
                cout << " " << i << "  ";
            else
                cout << i << "  ";
        }
        cout << "\n";
    }
    return 0;
}

OUTPUT

QUESTIONS

1) Is it wise to use pattern generating formulas? for eg, for putting value of i in last loop, i used the formula for the pattern \$1,2,4,7\$... as \$\frac{n*(n-1)}{2}+1\$. Is it efficient? I am asking this because this involves 4 operations in every iteration, which would make it prettu slow!

2) Is it possible to reduce no. of loops? is it better to reduce variables or reduce loops?

3) Can anyone show a recursive approach (or iterative)?

THANK YOU!