lightoj--1294--Positive Negative Sign（水题，规律）

Positive Negative Sign

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status

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 m 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 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

2

12 3

4 1

Sample Output

Case 1: 18

Case 2: 2

Source

Problem Setter: Jane Alam Jan

#include<stdio.h>
#include<string.h>
long long m,n;
int main()
{
	int t;
	scanf("%d",&t);
	int Case=1;
	while(t--)
	{
		long long sum=0;
		scanf("%lld%lld",&n,&m);
		printf("Case %d: ",Case++);
		sum=m*n/2;//就这麽一个简单的规律，超时好多次 
		printf("%lld
",sum);
	}
	return 0;
}