lightoj 1294 Positive Negative Sign 【规律】

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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 mintegers 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

题意：n 可以整除2*m,从1开始，每2*m 个数 减加求和

题解：有规律：每2*m个数的和为m*m,共有m*m*(n/(2*m))

代码：

#include <cstdio>int main(){    int t,k=1;    int n,m;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        printf("Case %d: %d\n",k++,n*m/2);    }    return 0;}