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  • light oj 1294

    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

    2

    12 3

    4 1

    Case 1: 18

    Case 2: 2

    #include<stdio.h>
    #include<string.h>
    #define LL long long
    int main()
    {
    	int t,k;
    	LL n,m,sum;
    	scanf("%d",&t);
    	k=1;
    	while(t--)
    	{
    		scanf("%lld%lld",&n,&m);
    		sum=(n/2)*m;
    		printf("Case %d: ",k++);
    		printf("%lld
    ",sum);
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/tonghao/p/4949669.html
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