codeforces - Tricky Sum（模拟）

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.

Sample test(s)

input

2
4
1000000000

output

-4
499999998352516354

Note

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

就是从1-n的和将2的次方都去掉。。。。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cmath>
using namespace std;
#define CRL(a) memset(a,0,sizeof(a))
typedef unsigned __int64 ll;
#define T 100005
#define mod 1000000007
ll table[35];
void paly_table()
{
for(int i=0;i<32;++i){
table[i] = 1<<i;//最多使用30
}
}
int main()
{
#ifdef zsc
freopen("input.txt","r",stdin);
#endif
int n;
ll sum,m;
scanf("%d",&n);
paly_table();
while(n--)
{
scanf("%I64d",&m);
sum = 0;
for(int i=0;i<=30;++i){
if(table[i]<=m){
sum += table[i];
}
else break;
}
printf("%I64d\n",m*(m+1)/2 - sum-sum);
}
return 0;
}