LightOJ 1294 - Positive Negative Sign【数学】

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

解题思路

有公式可以得出：将n除以二可得出有多少段，而m正好是每段的值，

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int cn=0;
    int t;
    scanf("%d",&t);
    while(t--)
    {
            long long n,m;
        scanf("%lld%lld",&n,&m);
        printf("Case %d: ",++cn);
        printf("%lld\n",n/2*m);
    }
    return 0;
}