【LightOJ】1294 - Positive Negative Sign

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1294 - Positive Negative Sign

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.

找规律问题，题目给出了 n 是2 * m 的倍数，所以题目很简单了。注意数据范围。

代码如下：

#include <cstdio>
int main()
{
	long long n,m;
	int u;
	int num = 1;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%lld %lld",&n,&m);
		printf ("Case %d: ",num++);
		long long ans;
		ans = m * n / 2;
		printf ("%lld\n",ans);
	}
	return 0;
}