1294

1294 - Positive Negative Sign

Time Limit: 2 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

12 3

4 1

Case 1: 18

Case 2: 2

题解：这个规律比较好推，直接每隔m一正一负，依次相减就是m了，因为整除2*m，就直接是m*n/2；因为没有加case，wa了一次，还是太不细心了啊；

代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define mem(x,y) memset(x,y,sizeof(x))
 7 using namespace std;
 8 typedef long long LL;
 9 const int INF=0x3f3f3f3f;
10 int main(){
11     int T,flot=0;
12     LL n,m;
13     scanf("%d",&T);
14     while(T--){
15         scanf("%lld%lld",&n,&m);
16         printf("Case %d: %lld
",++flot,m*(n/2));
17     }
18     return 0;
19 }