codeforces-598

最新推荐文章于 2020-11-15 12:36:45 发布

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本文介绍了一种名为TrickySum的算法实现，该算法要求计算从1到n的所有整数之和，但所有2的幂次方数需取负值。通过示例解释了计算过程，并提供了一段C++代码实现，用于批量处理多个n值。

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题目链接：点击打开链接

In this problem you are to calculate the sum of all integers from 1 to n, but you should take all powers of two with minus in the sum.

For example, for n = 4 the sum is equal to - 1 - 2 + 3 - 4 = - 4, because 1, 2 and 4 are 20, 21 and 22 respectively.

Calculate the answer for t values of n.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of values of n to be processed.

Each of next t lines contains a single integer n (1 ≤ n ≤ 109).

Output

Print the requested sum for each of t integers n given in the input.

   Examples
 

     input
   
2
4
1000000000

     output
   
-4
499999998352516354

Note

The answer for the first sample is explained in the statement.

水题啊，刚开始还要用快速幂写，真ZZ

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define LL long long
using namespace std;
LL n;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%I64d",&n);
		LL a=n*(n+1)/2;
		LL b=1;
		while(b<=n)
		{
			b<<=1;
		}
		printf("%I64d\n",a-(b-1)-(b-1));
	}
	return 0;
}