Codeforces 1077A - Frog Jumping
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.
Input:
The first line of the input contains one integer t (1≤t≤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≤a,b,k≤109) — the lengths of two types of jumps and the number of jumps, respectively.
Output:
Print t integers. The i-th integer should be the answer for the i-th query.
範例:
input:
1 | 6 |
output:
1 | 8 |
Note:
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 109−1+109−1+109−1=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.
題意:
青蛙第一次會向右跳a，第二次會向右跳b，問你跳k伺候青蛙位在多少？
思路:
兩次一組，往右跳a-b。因此偶數就直接算有幾組，奇數就再多加一次往右。