A - Tricky Sum CodeForces - 598A (推公式)
A - Tricky Sum CodeForces - 598A
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.
Examples
Input
2 4 1000000000
Output
-4 499999998352516354
Note
The answer for the first sample is explained in the statement.
题目大意:计算y= 1 - 2 + 3 - 4+5+6+...+i-2^k+.....+n;【1~n中2的k次方为负,其余为+】
可以化为y=1+2+3+.....+n-2*(2^0+2^1+...+2^k)=(1+n)*n/2-2*(2^(k+1)-1);
当时不太记得等比数列前n项和公式,懒得推,直接for求了,反正k很小
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <queue>
#include <set>
#include <map>
#define loopa(i,a,b,d) for(int i=a;i<b;i+=d)
#define loops(i,a,b,d) for(int i=a;i>b;i-=d)
typedef long long LL;
using namespace std;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
LL n;
scanf("%lld",&n);
LL sum=0;
for(LL i=1;i<=n;i*=2)
{
sum+=i;
}
sum*=2;
LL ans=n;
if(n%2==0)ans=ans/2*(ans+1);
else ans=(ans+1)/2*ans;
ans-=sum;
printf("%lld\n",ans);
}
return 0;
}