Tricky Sum
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.
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).
Print the requested sum for each of t integers n given in the input.
2 4 1000000000
-4 499999998352516354
The answer for the first sample is explained in the statement.
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
using namespace std;
struct P
{
int star;
int endd;
}num[900000];
bool cmp(P a,P b)
{
return a.endd<b.endd;
}
int main()
{
int n,k,p;
while(scanf("%d",&n)!=EOF)
{
k=1;
for(int i=0;i<n;i++)
{
scanf("%d%d",&num[i].star,&num[i].endd);
}
sort(num,num+n,cmp);
p=num[0].endd;
for(int i=1;i<n;i++)
{
if(num[i].star>p)
{
k++;
p=num[i].endd;
}
}
printf("%d\n",k);
}
return 0;
}