CodeForces - 598A Tricky Sum （快速幂）

CodeForces
- 598A
Tricky Sum

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

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 22respectively.

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.

//题意：

先输入一个T，表示有几组输入，再输入一个n，表示要计算n的和sum。

计算规则为，从1---n这n个数中任意一个数m，如果m这个数是2的次方的话，sum+=(-m)，否则sum+=m。

//思路：

以为数比较大，所以先求出总的和sum，再用快速幂，求出所有是2的次方的数的和num，最后求差即可sum=sum-2*num。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define IN __int64
#define N 10010
#define M 1000000007
#define PI acos(-1.0)
using namespace std;
ull kp(int n,int k)
{
ull s=1;
while(k)
{
if(k&1)
s*=n;
n*=n;
k/=2;
}
return s;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
IN n;
cin>>n;
IN sum=((n*(n+1))/2);
IN num=0;
int k=0;
while(1)
{
int m=kp(2,k);
if(m<=n)
{
num=num+(IN)m;
k++;
}
else
break;
}
sum=(sum-num*2);
printf("%I64d\n",sum);
}
return 0;
}