Lightoj1294——Positive Negative Sign（神坑）

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
Output for Sample Input
Case 1: 18
Case 2: 2

不说话，看代码

#include<stdio.h>#include<algorithm>#include<string.h>#include<iostream>#include<cmath>#define MAXN 10000using namespace std;int main(){    long long t,cnt=1,n,m,sum;    scanf("%I64d",&t);    while(t--)    {        scanf("%lld%lld",&n,&m);        sum=m*n/2;   //看到这个规律我是崩溃的        printf("Case %lld: %lld\n",cnt++,sum);  //以后TM再也不用I64d了！！    }    return 0;}