Kotlin Heroes: Practice 7 |
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Finished |
A frog is currently at the point $$$0$$$ on a coordinate axis $$$Ox$$$. It jumps by the following algorithm: the first jump is $$$a$$$ units to the right, the second jump is $$$b$$$ units to the left, the third jump is $$$a$$$ units to the right, the fourth jump is $$$b$$$ units to the left, and so on.
Formally:
Your task is to calculate the position of the frog after $$$k$$$ jumps.
But... One more thing. You are watching $$$t$$$ different frogs so you have to answer $$$t$$$ independent queries.
The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of queries.
Each of the next $$$t$$$ lines contain queries (one query per line).
The query is described as three space-separated integers $$$a, b, k$$$ ($$$1 \le a, b, k \le 10^9$$$) — the lengths of two types of jumps and the number of jumps, respectively.
Print $$$t$$$ integers. The $$$i$$$-th integer should be the answer for the $$$i$$$-th query.
6 5 2 3 100 1 4 1 10 5 1000000000 1 6 1 1 1000000000 1 1 999999999
8 198 -17 2999999997 0 1
In the first query frog jumps $$$5$$$ to the right, $$$2$$$ to the left and $$$5$$$ to the right so the answer is $$$5 - 2 + 5 = 8$$$.
In the second query frog jumps $$$100$$$ to the right, $$$1$$$ to the left, $$$100$$$ to the right and $$$1$$$ to the left so the answer is $$$100 - 1 + 100 - 1 = 198$$$.
In the third query the answer is $$$1 - 10 + 1 - 10 + 1 = -17$$$.
In the fourth query the answer is $$$10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997$$$.
In the fifth query all frog's jumps are neutralized by each other so the answer is $$$0$$$.
The sixth query is the same as the fifth but without the last jump so the answer is $$$1$$$.
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