Nezuko in the Clearing

Let’s strip away the anime fog and look at the actual machine.

Nezuko must move exactly $d$ times. That part is non-negotiable. The only thing we can optimize is how many times she rests. Therefore:

$\text{answer}=d+\text{minimum rests}.$

A rest does two things:

1. increases health by $1$;
2. resets the consecutive-move counter.

That second part is the entire problem.

Blocks of moves

Any valid plan can be seen as blocks of consecutive moves, separated by rests.

A block of length $k$ costs

$1+2+\cdots+k=\frac{k(k+1)}2$

health. Since health only decreases inside a block, it is enough to require that after the whole block the health is still positive.

So if block lengths are $a_1,a_2,\ldots,a_m$, total movement damage is

$\sum_{i=1}^{m} \frac{a_i(a_i+1)}2.$

The number of mandatory rests between those $m$ blocks is $m-1$.

Fix the maximum block length

Suppose no block is longer than $x$.

Then the minimum possible number of blocks is

$m=\left\lceil\frac d x\right\rceil,$

so we need at least

$m-1$

rests just to reset between blocks.

If

$d=qx+r \quad (0\le r<x),$

then using chunks of length $x$, plus one leftover chunk of length $r$, gives movement damage

$\text{damage}(x)=q\cdot\frac{x(x+1)}2+\frac{r(r+1)}2.$

For this $x$, the rests must cover two constraints:

• reset rests:

$\left\lceil\frac d x\right\rceil-1$

• healing rests, because final health must stay at least $1$:

$h+R-\text{damage}(x)\ge 1,$

so

$R\ge \text{damage}(x)-h+1.$

Therefore

$f(x)=\max\left(\left\lceil\frac d x\right\rceil-1,\ \text{damage}(x)-h+1\right).$

The answer is

$d+\min_{1\le x\le d} f(x).$

Why searching works

As $x$ grows:

• the reset-rest term decreases;
• the movement-damage term increases overall;
• their maximum forms a single valley around the balance point.

So we can ternary search over integer $x$. Since integer ceilings make the function jumpy, don’t be cute: ternary search until the interval is small, then brute force every remaining $x$.

For each $x$:

$q=\left\lfloor\frac d x\right\rfloor,\quad r=d\bmod x,$

$\text{damage}(x)=q\cdot\frac{x(x+1)}2+\frac{r(r+1)}2,$

$f(x)=\max\left(\frac{d+x-1}{x}-1,\ \text{damage}(x)-h+1\right).$

Then print $d+\min f(x)$.

Use long long; intermediate values fit. No Big Integer nonsense, thankfully.