CodeForces 598A （水题）

最新推荐文章于 2020-11-15 12:36:45 发布

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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，但若整数为2的幂则取负值。文章提供了完整的C++代码实现，包括输入多个n值并分别计算的过程。

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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
2
4
1000000000

output
-4
499999998352516354

Note

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

题意：给你n个数（1-n），如果为2^i,那么就把这个数为-1，求所有数之和。

先求出1-n的和，然后再减去2的次方就ok了。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		long long n;
		scanf("%lld",&n);
		long long sum=0;
		sum=(1+n)*n/2;
		long long i;
		long long ans=1;
		for(i=1;ans<=n;i++)
		{
			ans=ans*2;
			sum-=ans;
		}
		printf("%lld\n",sum);
	}
	return 0;
}