CodeForces - 598A Tricky Sum

CodeForces - 598A Tricky Sum
题目描述：
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.

题目大意：给出一个数n，从1运算到n，如果这个数是2的倍数，就为减，否则就为加。
思路：这道题。。。其实就是等差数列、等比数列求和。但是注意是减去两倍的等比数列。

#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>

using namespace std;
int t;
long long ans , n , i;

int main(){
    scanf("%d",&t);
    while(t--){
        i = 1;
        scanf("%lld",&n);
        ans = (n * (n + 1)) / 2;
        while(i <= n) i = i * 2;
        ans = ans - i * 2 + 2;
        printf("%lld\n",ans);
    }
    return 0;
}