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

241000000000

output

-4499999998352516354

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 1000000007ll 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;}