Educational Codeforces Round 1 (A)模拟

A. Tricky Sum

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

241000000000

output

-4499999998352516354

Note

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

题意：1到n中，2的指数次方的数字为负数。如2^0,2^1,2^2,2^3。。。。。。剩下的都是正数。把正数-负数，不可以直接模拟，复杂度O(N),一直乘2，复杂度为log（N）.

#include<iostream>#include<cstdio>using namespace std;#define LL long longint main(){    int t;LL n;    scanf("%d",&t);    while(t--)    {        scanf("%lld",&n);        LL sum1=0,sum2;        LL tt=1;       while(n>=tt)       {        sum1+=tt;        tt<<=1;       }       sum2=n*(n+1)/2-sum1;       printf("%lld\n",sum2-sum1);    }    return 0;}