Fill an array diagonally - c++

Question

I have following program. with Input 3 5
3 rows
5 growth of numbers
The output should be:
1 2 4 7 10
3 5 8 11 13
6 9 12 14 15
But my program gives:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Here is what I have tried so far
int main() {
int n, m, c = 0;
cin >> n >> m;
int a[n][m];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
a[i][j] = ++c;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++)
cout << setw(4) << a[i][j];
cout << endl;
}
}
What I am doing wrong or missing?
About the spaces: Can't find reason for such behavior(first spaces are ignored), displayed on screenshot. Tried to run in different IDE's with different compilers and had such problem only in testing system.

Answer 1

Hi try to use tab instead.
#include <iostream>
using namespace std;
int main() {
int n, m, c = 0;
cin >> n >> m;
int *a = new int[n * m];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
a[i * n + j] = ++c;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++)
cout << "\t" << a[i * n + j];
cout << endl;
}
delete[] a;
return 0;
}

Answer 2

Can't remember how I solved this problem in secondary school, but with n less than m, the following code works:
#include <iostream>
using namespace std;
void nextij(long n,long m,long& i,long& j) {
if (i==n-1) { //bottom row
if (j<n-1) { //in the left square
j = i+j+1;
i = 0;
}
else { //out of the left square
i = j-(n-1)+1;
j = m-1;
}
}
else { //other rows
if (j==0) { //left most column
j = i+1;
i = 0;
}
else { //other columns
i++;
j--;
}
}
}
int main() {
long n = 3;
long m = 5;
long a[3][5];
long i = 0;
long j = 0;
long c = 1;
while (c<=n*m) {
a[i][j] = c;
nextij(n,m,i,j);
c++;
}
for (i=0; i<n; i++) {
for (j=0; j<m; j++)
cout <<a[i][j] <<" ";
cout <<endl;
}
}
/*
output:
1 2 4 7 10
3 5 8 11 13
6 9 12 14 15
*/