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.
Examples
input
Copy
2 4 1000000000
output
Copy
-4 499999998352516354
Note
The answer for the first sample is explained in the statement.
代码：
#include<set>
#include<map>
#include<cmath>
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
long long n;
long long sum;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
long long sum1=n*(n+1)/2;
long long sum2=0;
for(int i=2;i<=n;i=i*2)
sum2+=i;
printf("%lld %lld\n",sum1,sum2);
sum=sum1-2*sum2-2;
printf("%lld\n",sum);
}
return 0;
}