J - Positive Negative Sign

最新推荐文章于 2020-01-12 19:54:58 发布

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本文介绍了一种针对特定整数序列的高效求和算法。该算法通过将序列中的整数按规律分配正负号，然后计算总和。输入包括一系列测试用例，每个用例由两个整数n和m组成，输出则是按照指定规则计算得到的序列总和。

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

int n,m; NND坑死我了

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