Coderforce 598A Tricky Sum

Tricky Sum 

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

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

241000000000

Output

-4499999998352516354

//求1到n的和，但是要减去1到n中所有的2的倍数

AC代码：

#include<stdio.h>#include<math.h>using namespace std;int main(){int t;long long n;long long s1=0;long long s2=0;scanf("%d",&t);while(t--){int k=0;scanf("%lld",&n);for(int i=0; ;i++){if(pow(2,i)==n){k=i; break;}    else if(pow(2,i)>n){ k=i-1; break;}       else        continue;}s1=n*(1+n)/2;s2=pow(2,k+1)-1;printf("%lld\n",s1-2*s2);}return 0; }