1294 - Positive Negative Sign

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Time Limit:2 second(s)

Memory Limit:32 MB

Given two integers:nandmandnis divisible by2m, you have to write down the firstnnatural numbers in the following form. At first take firstmintegers and make their sign negative, then take nextmintegers and make their sign positive, the nextmintegers should have negative signs and continue this procedure until all thenintegers have been assigned a sign. For example, letnbe12andmbe3. Then we have

-1 -2 -3+4 +5 +6-7 -8 -9+10 +11 +12

Ifn = 4andm = 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 integerT (≤ 10000), denoting the number of test cases.

Each case starts with a line containing two integers:nandm (2 ≤ n ≤ 10⁹, 1 ≤ m). And you can assume thatnis divisible by2*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

題解：這個(gè)規(guī)律比較好推，直接每隔m一正一負(fù)，依次相減就是m了，因?yàn)檎?*m，就直接是m*n/2；因?yàn)闆]有加case，wa了一次，還是太不細(xì)心了啊；

代碼：

 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 }

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