【codeforces】(等差数列及等比数列求和）

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.

思路：先求从1到n的等差数列的和（d=1)sum1,在求从1到m等比数列的和（q=1)(m<=2^n)sum2;

答案即为sum1-2*sum2;

#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
int i;
__int64 m,sum,sum1=0,sum2=0;
scanf("%I64d",&m);
sum1=(m+1)*m/2;
int l;
for(i=0;;i++)
{
if(pow(2,i)==m)
{
l=i;
break;
}
if(pow(2,i)>m)
{
l=i-1;
break;
}
}
sum2=pow(2,l+1)-1;
sum=sum1-2*sum2;
printf("%I64d\n",sum);
}
}