LightOJ 1294 - Positive Negative Sign【数学】

浅浅的花香味﹌ 2022-08-18 05:56 150阅读 0赞

1294 - Positive Negative Sign

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   <td>Time Limit: <span style="color:rgb(180,95,4)">2 second(s)</span></td> 
   <td>Memory Limit: <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>Sample Input</h1> </td> 
   <td> <h1>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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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;
    }