Tricky Sum(等比数列)

Tricky Sum
Crawling in process...Crawling failedTime
Limit:1000MS    Memory Limit:262144KB    
64bit IO Format:%I64d & %I64u
SubmitStatus

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 are20,
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 ofn to be processed.

Each of next t lines contains a single integern (1 ≤ n ≤ 109).

Output

Print the requested sum for each of t integersn given in the input.

Sample Input

Input

2
4
1000000000

Output

-4
499999998352516354

Sample Output

302000

 

Hint

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

求前n个数的和，1-------n;(其中等于2^m(m=0、1、2、。。。。。)都为负数，其余数都为负数。

2的所有次方和：s=2^(t+1)-1（等比数列求和且2^t<n）

    h=1+2+3+4+……+n=n(n+1)/2

最后结果：

ans=h-2*t;

My  solution:

/*2016.3.12*/

<pre name="code" class="cpp">#include<stdio.h>
long long n;
long long ans,su;
void  jie()
{
int i,t1=0;
long long time=2;
ans=1;
while(ans<=n)
{
t1++;
ans*=2;
}
i=t1;
ans=1;
while(i)
{
if(i%2)
ans*=time;
i/=2;
time*=time;
}
return ;
}
int main()
{
long long i;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%I64d",&n);
jie();
if(n%2==0)
su=(n/2)*(n+1);
else
su=((n+1)/2)*n;
su=su-2*(ans-1);
printf("%I64d\n",su);
}

return 0;
}