在C中以一定条件打印直角三角形

Question

Enter a certain number, and that number is a condition which determines the number of chars in a single row.Let's say the number is 3 In the first row there is only 1 char. In second row there is a condition. a+1 Where a is the number that we entered In third row is 2a+1 Fourth 3a+1 And so on... Example: Number that we entered is 3.

a (1)
aaaa (3+1)
aaaaaaa (2*3+1)

Here is what i've came up. I have trouble with implementing that condition.

#include<stdio.h>
main()
{
int i,j,n;

printf("Enter the numbers of rows:");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=i;j++)
printf("a");
printf("\n");
}
getch();
} 

Answer 1

Just start iterating inner for loop starting from n till i * n as below,

for(i=1;i<=n;i++)
{
  for(j=n;j<=i*n;j++)
  printf("a");
  printf("\n");
}

Here is the demo

Answer 2

Suggestion: get used to start counting at 0

for(i=1;i<=n;i++) // could be for (i = 0; i < n; i++)
{
for(j=1;j<=i;j++) // could be for (j = 0; j < i; j++)

---

You have to have a multiplication by 3 "somewhere". Try and find the right place and what to multiply by 3.

Answer 3

for(i=0;i<n;i++){
    for(j=0;j<n*i+1;j++)
        printf("a");
    printf("\n");
}