Codeforces--598A--Tricky Sum（数学）



Tricky Sum

Tricky SumCrawling in process...
Crawling failed
Time Limit:1000MS    
Memory Limit:262144KB    
64bit IO Format:%I64d & %I64u

Submit
Status Practice CodeForces 598A

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

Sample Output

Hint

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

1--n全部加起来，如果是2的次幂  数字就为负的，求最后的和，先利用等差序列公式（n+1）*n/2，然后扣除2的次幂

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		__int64 n;
		cin>>n;
		__int64 ans=(n+1)*n/2-2;
		for(int i=1;pow(2,i)<=n;i++)
		ans-=pow(2,i+1);
		cout<<ans<<endl;
	}
	return 0;
}