Yang Hui's triangle C language

describe:

Print the Yanghui triangle

Enter description:

The first line contains an integer number n. (1≤n≤30)

Output description:

Contains n lines, which are the first n lines of Yanghui's triangle.

E.g:

Input: 6

output:

    1
    1    1
    1    2    1
    1    3    3    1  
    1    4    6    4    1
    1    5   10   10    5    1

Ideas:

After many students wrote 2 layers of for loops, there is no more. The main reason is that it is more difficult to print 1 in the outermost layer. After solving the outermost 1, the inner number can be obtained through the properties of the Yanghui triangle.

We still have to use two-dimensional arrays to solve the problem by finding rules.

We start at line 0. 1 in row 0 corresponds to i = 0, j = 0; 1 in row 1 corresponds to i = 1, j = 0; i = 1, j = 1; 1 in row 2 corresponds to i = 2, j = 0; i = 2, j = 2; In this way, we find that 1 is printed when i == j, and, no matter how i changes, 1 is also printed when j = 0, so  

if((i == j) || j == 0)
			{
				arr[i][j] = 1;
				printf("%d ", arr[i][j]);
			}

After solving the 1 in the outer layer, the numbers inside are solved very well. For example, arr[2][1] (that is, the number 2) = arr[1][0] + arr[1][1], or find the law and find:

arr[i][j] = arr[i - 1][j - 1] + arr[i - 1][j];

Then the complete code is as follows:

#include<stdio.h>

int main(void)
{
    int n = 0;
    int i = 0, j = 0;
    int arr[30][30] = { 0 };
    scanf("%d", &n);
    for(i = 0; i < n; i++)
    {
        for(j = 0; j <= i; j++)
        {
            if(j == 0 || j == i)
            {
                arr[i][j] = 1;
                printf("%d", arr[i][j]);
            }
            else
            {
                arr[i][j] = arr[i - 1][j - 1] + arr[i - 1][j];
                printf("%d", arr[i][j]);
            }
        }
        printf("\n");
    }
    
    return 0;
}

Some students may taste the sweetness of one-dimensional arrays on certain topics, and want to use one-dimensional arrays to solve this problem. However, the courage is commendable (0-0)!