【codeforces】数学

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本文介绍了一个算法问题，即计算从1到n的所有整数之和，但所有2的幂次项需要取负值。通过预先计算2的幂次并存储在一个数组中，再结合等差数列求和公式，该问题可以高效解决。

A - A 使用long long

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

Submit Status Practice CodeForces 598A

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

2
4
1000000000

Output

-4
499999998352516354

Hint

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

#include<cstdio>
#include<cmath>
#include<algorithm>
#define LL long long
using  namespace std;
LL a[50];
void init(){
	for(int i=0;i<=32;i++){
		a[i]=pow(2,i);
	}
}
int main(){
	LL n,t;
	scanf("%lld",&t);
	init();
	while(t--){
		LL sum,num=0;
		scanf("%lld",&n);
		sum=n*(n+1)/2;                  //等差数列
	//	printf("%lld--\n",sum);
		for(int l=0;l<=32&&a[l]<=n;l++){
			num=num+a[l];
		}
		printf("%lld\n",sum-num*2);
	}
	return 0;
}