I want to print the following pattern - c++

Question

1
3 2
6 5 4
10 9 8 7
I want to print the following pattern. I have tried very hard but couldn't make the code for it. I have tried everything which came up to my mind.
#include<iostream>
using namespace std;
int main() {
int i, j, n;
cin >> n;
int k = 0;
for (i = 1;i <= n; i++) {
for (j = 1; j <= i; j++) {
k++;
printf("%d ", k);
}
printf("\n");
}
}
the other code which i tried is this.
#include<iostream>
using namespace std;
int main() {
int i, j, n;
cin >> n;
int k = 0;
for (i = 1; i <= n; i++) {
for (j = i; j >= 1; j--) {
k++;
printf("%d ",j);
}
printf("\n");
}
}

Answer 1

#include <iostream>
#include <stack>
using namespace std;
int main()
{
int previousRow = 0;
for(int row = 1; row <= 4; row++)
{
int rowTracker = row;
for(int col = 0; col < row; col++)
{
cout<<rowTracker - col + previousRow<<" ";
}
previousRow += row;
cout<<endl;
}
return 0;
}

Answer 2

#include<iostream>
void printPattern(unsigned numlevels)
{
unsigned last_num = 1;
for(unsigned i = 0; i < numlevels; ++i)
{
unsigned next_num = i + last_num;
for(unsigned j = next_num; j >= last_num; --j)
{
std::cout << j << ' ';
}
std::cout << '\n';
last_num = next_num + 1;
}
}
int main()
{
unsigned n;
std::cin >> n;
printPattern(n);
return 0;
}

Answer 3

You may also use a stack to implement this. Here is a working answer:
#include <iostream>
#include <stack>
using namespace std;
int main() {
int i, j, n;
stack<int> st;
cin >> n;
int k = 0;
for(i = 1;i <= n; i++) {
for(j = 1; j <= i; j++) {
k++;
st.push(k);
}
while(!st.empty()){
printf("%d ", st.top());
st.pop();
}
printf("\n");
}
}
Hope it helps!

Answer 4

Before reading the code blow, you should really try to do it yourself. This problem is obviously for practice and to develop the programming muscle. Just getting the answer is not going to help.
The issue with your code is that for each row, the range you want to print is not being determined correctly. You should first find the range and then print the numbers. Ther can be multiple approaches to it. Below is one of them.
for(i=1;i<=n;i++){
int max = i*(i+1)/2;
int min = i*(i-1)/2 + 1;
for(j=max;j>=min;j--){
printf("%d ",j);
}
printf("\n");
}

Answer 5

Here is a simple method
int main(int argc, char* argv[])
{
int n = 4; // suppose print 4 lines
for (int i = 1; i <= n; ++i)
{
int i0 = (i + 1) * i / 2; // first number of line i
for (int j = 0; j < i; j++)
cout << i0 - j << " ";
cout << endl;
}
return 0;
}

Answer 6

thanx everyone for your responses. I was able to do it on my own. Below is what I did. if there is any correction let me know
#include<iostream>
using namespace std;
int main(){
int i,j,n,temp;
cin>>n;
int k=0;
for(i=1;i<=n;i++){
k=k+i,temp=k;
for(j=1;j<=i;j++){
cout<<temp<<+" ";
temp--;
}
cout<<("\n");
}
}