Easy Problem VI

SCU-4486
Time Limit: 1000 MS Memory Limit: 131072 K

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 2^0, 2^1 and 2^2 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≤10^9).

Output

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

Sample Input

2
4
1000000000

Sample Output

-4
499999998352516354
注释：注意不超时就行了

#include<stdio.h>#include<string.h>long long v[1000];void  ff(){    v[1]=1;       for(int i=2; v[i]<=1000000000; i++)    {//1<=2^n<=1000000000        v[i]=v[i-1]*2;    }}int main(){    int T;    ff();    scanf("%d",&T);    while(T--)    {        long long n;        int k;        long long sum=0,ans=0;        scanf("%lld",&n);        sum=n*(n+1)/2;//等差公式不容易超时        for(int i=1;v[i]<=n;i++)            ans+=2*v[i];        printf("%lld\n",sum-ans);    }    return 0;}