CodeForces 598A Tricky Sum

最新推荐文章于 2018-03-20 23:28:15 发布

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最新推荐文章于 2018-03-20 23:28:15 发布

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本文介绍了一个计算从1到n的所有整数之和的问题，但所有2的幂次方数需取负值。通过直接计算的方法给出了针对多个n值求解的程序实现。

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Q - Q

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Description

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.

Sample Input

Input

2
4
1000000000

Output

-4
499999998352516354

Hint

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

没想到直接计算竟然AC了，本来以为会有什么技巧呢，真水。。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
	__int64 t,i,n,sum;
	scanf("%I64d",&t);
	while(t--)
	{
		sum=0;
		scanf("%I64d",&n);
		sum=n*(n+1)/2;
		for(i=1;i<=n;i=i*2)
		{
			sum=sum-2*i;
		}
		printf("%I64d\n",sum);
	}
	return 0;
}