LightOJ 1294:Positive Negative Sign

B - Positive Negative Sign

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1294

Appoint description: jehad  (2012-06-14)System Crawler  (2015-11-04)

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 nintegers 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

需注意的是10的九次方*10的九次方。。。。。long long

AC-code:

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