考えたこと
\begin{aligned}
\sum_{i=1}^{k} = \frac{k(k+1)}{2}
\end{aligned}
だから，\sum_{i=1}^{k} = \frac{k(k+1)}{2}
\end{aligned}
\begin{aligned}
\frac{k(k+1)}{2} \geq N
\end{aligned}
を満たす最小の正整数$k$を求めれば良い．\frac{k(k+1)}{2} \geq N
\end{aligned}
２次方程式
\begin{aligned}
& \frac{k(k+1)}{2} = N \\
&\Leftrightarrow k^{2} + k - 2N = 0
\end{aligned}
の解は& \frac{k(k+1)}{2} = N \\
&\Leftrightarrow k^{2} + k - 2N = 0
\end{aligned}
\begin{aligned}
k=\frac{-1\pm \sqrt{1 + 8N}}{2}
\end{aligned}
であるが，$k > 0$となるのはk=\frac{-1\pm \sqrt{1 + 8N}}{2}
\end{aligned}
\begin{aligned}
k=\frac{-1 + \sqrt{1 + 8N}}{2}
\end{aligned}
である．よって，$\lceil k \rceil=$k=\frac{-1 + \sqrt{1 + 8N}}{2}
\end{aligned}
math.ceil(k)が答え．別解
Editorial - AtCoder Beginner Contest 206（Sponsored by Panasonic）二分探索は覚えておきたい．
