lightij-1294 - Positive Negative Sign【思维】

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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.

一道水题，没看出来规律，就想了那么长时间；ZZ

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long
using namespace std;
LL n,m;
int main()
{
  int t,text=0;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%lld %lld",&n,&m);
    LL ans=m*(n/2);
    printf("Case %d: %lld\n",++text,ans);
  }
}