LIGHTOJ-1294 - Positive Negative Sign

淡淡的烟草味﹌ 2022-08-18 02:10 136阅读 0赞

1294 - Positive Negative Sign

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   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Time Limit:&nbsp;<span style="color:rgb(180,95,4)">2 second(s)</span></td> 
   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Memory Limit:&nbsp;<span style="color:rgb(180,95,4)">32 MB</span></td> 
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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.

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   <td> <h1 style="color:green">Sample Input</h1> </td> 
   <td> <h1 style="color:green">Output for Sample Input</h1> </td> 
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   <td> <p>2</p> <p>12 3</p> <p>4 1</p> </td> 
   <td> <p>Case 1: 18</p> <p>Case 2: 2</p> </td> 
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#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int main()
    {
    	long long a,b,c,i,n,cot=1;
    	scanf("%lld",&n);
    	while(n--)
    	{
    		scanf("%lld%lld",&a,&b);
    		printf("Case %lld: %lld\n",cot++,a*b/2);
    		
    	}
    	return 0;
    }