LightOJ

Positive Negative Sign

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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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 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

2

12 3

4 1

Sample Output

Case 1: 18

Case 2: 2

Source

Problem Setter: Jane Alam Jan

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有判断条件时考虑极限情况;   利用取余判是否可整除, 注意根据题意加判断条件 ;

 

#include <cstdio>
#define LL long long
int main()
{
    int t, Q=1;
    scanf("%d", &t);
    while(t--)
    {
        LL n, m; 
        scanf("%lld%lld", &n, &m);
        LL q=m*m;
        LL gs=n/2/m;
        printf("Case %d: %lld
", Q++, gs*q);
    }
    return 0;
}