For Python IO we will start with a pretty basic project that reads in a number n and gives you Pascal's triangle with that number of rows. The code is listed below:
pascal.py

`import sysdef pascal(n):   row = [1]   k = [0]   for x in range(max(n,0)):      print row      row=[l+r for l,r in zip(row+k,k+row)]   return n>=1num = int(sys.stdin.readline())pascal(num)`

Above is the code to take in a number n and generate a Pascal's triangle with n rows.

Sample test cases:

`Test 0Input:1Expected Output:[1]Test 1Input:3Expected Output:[1][1, 1][1, 2, 1]Test 2Input:10Expected Output:[1][1, 1][1, 2, 1][1, 3, 3, 1][1, 4, 6, 4, 1][1, 5, 10, 10, 5, 1][1, 6, 15, 20, 15, 6, 1][1, 7, 21, 35, 35, 21, 7, 1][1, 8, 28, 56, 70, 56, 28, 8, 1][1, 9, 36, 84, 126, 126, 84, 36, 9, 1]Test 3Input:18Expected Output:[1][1, 1][1, 2, 1][1, 3, 3, 1][1, 4, 6, 4, 1][1, 5, 10, 10, 5, 1][1, 6, 15, 20, 15, 6, 1][1, 7, 21, 35, 35, 21, 7, 1][1, 8, 28, 56, 70, 56, 28, 8, 1][1, 9, 36, 84, 126, 126, 84, 36, 9, 1][1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1][1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1][1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1][1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1][1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1][1, 15, 105, 455, 1365, 3003, 5005, 6435, 6435, 5005, 3003, 1365, 455, 105, 15, 1][1, 16, 120, 560, 1820, 4368, 8008, 11440, 12870, 11440, 8008, 4368, 1820, 560, 120, 16, 1][1, 17, 136, 680, 2380, 6188, 12376, 19448, 24310, 24310, 19448, 12376, 6188, 2380, 680, 136, 17, 1]`