CareerCup

2 cuts........1 2 and 4
1st day..giv 1
2nd day giv 2 n tek 1
3rd day giv 1
4th day giv 4 n tek 1 n 2..
carry n.

this is a binary problem. sg has the right answer--you use powers of 2 (in this case, 1, 2, 4). if, for example, you had to divide 15 into as few segments as possible but still get all possible integers, you'd do 1, 2, 4, 8.

alternatively, if it was a ternary problem (for example, you want to weigh things using a scale, so the 3 options are 2 sides of the scale or not using a block) then you'd do powers of 3. so if you wanted to weigh all possible integer weights between 1 and 40, you'd do 1, 3, 9, 27.

3 cuts into 1 segment each
1st: cut into 4&3.
2nd: put 3 onto 4. cut at 2. Now we have 3 two-segment and 1 one-segment.
3nd: put 3 two-segment onto a stack, cut at 1.

1st - give 1
2nd day - yesterday's 1 + give 1 = 2
3rd day - 2 + 1 = 3
4th - 3 + 1 = 4
5th - 4 + 1 = 5
6th - 5 + 1 = 6
7th - 6 + 1 = 7