lightoj 1294 - Positive Negative Sign

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 mintegers 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

 

代码：

#include<stdio.h>#include<string.h>int main(){long long T,k=1;long long n,m;long long sum;scanf("%lld",&T);while(T--){scanf("%lld%lld",&n,&m);sum=(n*m)/2;printf("Case %lld: ",k++);printf("%lld\n",sum);}return 0;}