CF 598A Tricky Sum【规律】

A. Tricky Sum
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 12 and 4 are 2021 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.


求一下1到n的和  再减去2的倍数。即为所得

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <cmath>  
#include <cstdlib>   
#include <queue>  
#include <stack> 
#include <algorithm>
#define Wi(a) while((a)--) 
#define Si(a) scanf("%d", &a)  
#define Si64(a) scanf("%I64d", &a)
#define Pi(a) printf("%d\n", (a)) 
#define Pl(a) printf("%I64d\n", (a)) 
#define CLR(a, b) memset(a, (b), sizeof(a)) 
#define INF 0x3f3f3f3f   
#define LL long long 
using namespace std;
int main()
{
	int t;   Si(t);
	Wi(t)
	{
		__int64 n, ans;  Si64(n);
		ans = n*(n+1)/2;
		for(int i = 0; pow(2, i) <= n; ++i)
		{
			ans -= (2*(pow(2, i)));
		}
		Pl(ans);
	}
	return 0;
}



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