Positive Negative Sign<大水题>

最新推荐文章于 2020-11-20 17:38:55 发布

原创最新推荐文章于 2020-11-20 17:38:55 发布 · 483 阅读

0 ·

CC 4.0 BY-SA版权

水题同时被 2 个专栏收录

43 篇文章

订阅专栏

规律

9 篇文章

订阅专栏

本文介绍了一种符号化求和算法，该算法接受两个整数n和m作为输入，并将前n个自然数按特定规则分配正负号，然后计算这些数的总和。具体规则为：首先取前m个整数并赋予负号，接下来的m个整数赋予正号，依此类推。文章还提供了一个示例程序，用于演示如何根据输入的n和m来计算最终的符号化求和结果。

Description

Given two integers: n and m and n is divisible by 2m, you have to write down the first n natural numbers in the following form. At first take first mintegers 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 the n integers have been assigned a sign. For example, let n be 12 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 by 2*m.

Output

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

Sample Input

12 3

4 1

Sample Output

Case 1: 18

Case 2: 2

#include<cstdio>
int main()
{
	long long t,cut=0;
	scanf("%lld",&t);
	while(t--)
	{
		cut++;
		long long n,m;
		scanf("%lld%lld",&n,&m);
		printf("Case %lld: %lld\n",cut,n/2*m);
	}
	return 0;
}