codeforces 598A Tricky Sum-优快云博客

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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.

这个题主要是时间限制，刚开始我想从1到n循环一遍，结果超时，后来还是我的一位哥们提醒了我，先计算1到n，这个可以直接用公式，其他的我就不多说了。

#include<bits/stdc++.h>
#include<cmath>
using namespace std;
typedef long long ll;
int main()
{
    int w[50000],k=0;
    for(int i=1;i<=1000000000;i*=2)
    {
        w[k++]=i;
    }
    int t;
    ll n;
    cin>>t;
    while(t--)
    {
        ll sum;
        cin>>n;
        if(n%2==0)
            sum=n/2*(n+1);
        else
            sum=(n+1)/2*n;
        int i=0;
        while(w[i]<=n)
        {
            sum-=2*w[i];
            i++;
        }
        cout<<sum<<endl;
    }
    return 0;
}