light--oj--1294-- Positive Negative Sign（数学规律）

Given two integers: n and m and n is divisible by2m, you have to write down the first n natural numbers in the following form. At first take firstm integers and make their sign negative, then take next m integers and make their sign positive, the next m integers should have negative signs and continue this procedure until all then integers have been assigned a sign. For example, let n be12 and m be 3. Then we have

-1 -2 -3 +4 +5 +6 -7 -8 -9 +10 +11 +12

If n = 4 and m = 1, then we have

-1 +2 -3 +4

Now your task is to find the summation of the numbers considering their signs.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing two integers: n and m (2 ≤ n ≤ 10⁹, 1 ≤ m). And you can assume that n is divisible by2*m.

Output

For each case, print the case number and the summation.

Sample Input

2

12 3

4 1

Sample Output

Case 1: 18

Case 2: 2

题意：给两个数，n和m，其中n是2m的倍数，按题目描述的规律进行运算。

思路：推导出数学公式，并求得答案。

#include<stdio.h>int main(){int T,cnt=0;scanf("%d",&T);while(T--){long long m;long long n;scanf("%lld%lld",&n,&m);long long ans=m*n/2;printf("Case %d: %lld\n",++cnt,ans);}return 0;}