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No.2067 ±2^k operations

レベル : / AI推定難易度 : 2050 / 実行時間制限 : 1ケース 2.000秒 / メモリ制限 : 512 MB / 標準ジャッジ問題
タグ : （AC するまで非表示） / 解いたユーザー数 28
作問者 : とりゐ / テスター : 👑 qand ygussany karinohito
お気に入りにしたユーザー ProblemId : 8443 / 自分の提出
問題文最終更新日: 2022-09-02 02:08:55
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yukicoder contest 359 (順位表) の他の問題:

Problem Statement

Let $n$ be a positive integer. Starting from $X=n$, let $f(n)$ be the minimum number of operations required to make $X=0$ by repeating the following operation.

  • Choose a non-negative integer $k$ and one of the operators $+,-$.
    • If the operator is $+$, replace $X$ with $X+2^k$.
    • If the operator is $-$, replace $X$ with $X-2^k$.
Given a positive integer $N$, find $\displaystyle \sum_{n=1}^N f(n)$.

You are given $T$ test cases.

Input

$T$
$\mathrm{case}_1$
$\vdots$
$\mathrm{case}_T$
Each case is given in the following format. 
$N$

  • $1\leq T\leq 10^4$
  • $1\leq N\leq 10^{17}$
  • All inputs are integers.

Output

Print $T$ lines.

Sample

Sample 1
Input
3
3
10
2022
Output
4
16
8342

$f(1),f(2),f(3)$ are as follows.

  • When $n=1$, choosing $(0,-)$ in the $1$-st operation makes $X=0$. Thus, $f(1)=1$.
  • When $n=2$, choosing $(1,-)$ in the $1$-st operation makes $X=0$. Thus, $f(2)=1$.
  • When $n=3$, choosing $(2,-)$ in the $1$-st operation makes $X=-1$, and choosing $(0,+)$ in the $2$-nd operation makes $X=0$. Since it is impossible to reach $0$ in $1$ operation, $f(3)=2$.
Therefore, the answer to the $1$-st test case is $f(1)+f(2)+f(3)=1+1+2=4$.

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