Educational Codeforces Round 1 A. Tricky Sum 暴力

A. Tricky Sum

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/598/problem/A

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

2

4

1000000000

Sample Output

-4

499999998352516354

HINT

 

题意

让你输出前n个数的和，但是如果遇到了2的次方，就是减去他

题解：

正常做法，就直接O(1)得到前n个数的和，然后再log判断前n个数的2的倍数的数是哪些就好了。

但是我为什么脑抽了去写二分。。。

代码

 

#include<iostream>
#include<vector>
#include<stdio.h>
#include<algorithm>
using namespace std;

vector<long long>Q;
vector<long long>Sum;
int main()
{
    long long sum = 1;
    while(sum < 1e9+1)
    {
        Q.push_back(sum);
        sum*=2;
    }
    long long Sum1 = 0;
    for(int i=0;i<Q.size();i++)
    {
        Sum1 += Q[i];
        Sum.push_back(Sum1);
    }
    int t;scanf("%d",&t);
    while(t--)
    {
        long long n;cin>>n;
        long long ans = (1+n)*n/2;
        long long p = upper_bound(Q.begin(),Q.end(),n)-Q.begin();
        p--;
        ans -= 2*Sum[p];
        cout<<ans<<endl;
    }
}