1294 - Positive Negative Sign（规律）

1294 - Positive Negative Sign

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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
2 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：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #define mem(x,y) memset(x,y,sizeof(x))
 using namespace std;
 typedef long long LL;
 const int INF=0x3f3f3f3f;
 int main(){
     int T,flot=;
     LL n,m;
     scanf("%d",&T);
     while(T--){
         scanf("%lld%lld",&n,&m);
         printf("Case %d: %lld\n",++flot,m*(n/));
     }
     return ;
 }