###### 题目描述 对于给定的$a,b,k$，初始状态$x=0$，第$i$次操作满足以下规则： - $i$为奇数，$x=x+a$ - $i$为偶数，$x=x-b$ 求$k$次操作后$x$的值 ###### 输入格式 **题目有多组数据** 第一行一个整数$t(1\leq t\leq1000)$表示数据组数 接下来$t$行，每行三个整数$a,b,k(1\leq a,b,k\leq10^9)$表示每组数据 ###### 输出格式 输出共$t$行，第$i$行表示第$i$组数据的答案
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: - if the frog has jumped an even number of times (before the current jump), it jumps from its current position $ x $ to position $ x+a $ ; - otherwise it jumps from its current position $ x $ to position $ x-b $ . 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 $ .