A partial answer to question (the last part) is as follows:
(Assuming the two statements in the question are true)
If instead of printing number of split parts, if you print no. of occurrences of \n
j times, you'll get (simply doing a -1):
(1283168,1) (175,2) (17,3) (12,4) (10,5) (8,6) (7,7) (6,8) (5,9) (5,10) (4,11) (4,12) (3,13) (3,14) (3,15) (3,16) (2,17) (2,18) (2,19) (2,20) (2,21) (2,22) (2,23) (2,24) (2,25) (1,26) (1,27) (1,28) (1,29) (1,30) (1,31) (1,32) (1,33) (1,34) (1,35) (1,36) (1,37) (1,38) (1,39) (1,40) (1,41) (1,42) (1,43) (1,44) (1,45) (1,46) (1,47) (1,48) (1,49) (1,50)
Note that for j>3, product of both numbers is <=50, which is your maximum. What this means is that there is a place with 50 consecutive \n
characters and all the hits you are getting from 4 to 49 are actually part of the same.
However for 3, the maximum multiple of 3 less than 50 is 48 which gives 16 while you have 17 occurrences here. So there is an extra \n\n\n
somewhere with non-\n
character on both its 'sides'.
Now for 2 (\n\n
), we can subtract 25 (coming from the 50 \n
s) and 1 (coming from the separate \n\n\n
) to obtain 175-26 = 149.
Counting for the discrepancy, we should sum (2-1)*149 + (3-1)*1 + (50-1)*1, the -1 coming because first \n
in each of these is accounted for in the Scanner counting. This sum is 200.