Five. (20 points) Let $a$ be a given positive constant, and $n$ be a given natural number. If there exists a sequence of positive real numbers $\left\{x_{i}\right\}(i=1$, $2, \cdots, n)$ satisfying
(1) $\sum_{i=1}^{n} x_{i}=a$;
(2) $c \geqslant \frac{x_{j}}{1+x_{1}+x_{2}+\cdots+x_{j}}, 1 \leqslant j \leqslant n$,

find the minimum value of the real number $c$, and the corresponding $x_{i}$ at this time.
For $\left\{x_{i}\right\}$ that satisfy the requirements, we have
$$
1-c \leqslant \frac{1+x_{1}+\cdots+x_{j-1}}{1+x_{1}+\cdots+x_{j}}, 1 \leqslant j \leqslant n \text {. }
$$

Multiplying the above $n$ inequalities, we get
$(1-c)^{n}$
$\leqslant \frac{1}{1+x_{1}} \cdot \frac{1+x_{1}}{1+x_{1}+x_{2}} \cdots \cdot \frac{1+x_{1}+\cdots+x_{n-1}}{1+x_{1}+\cdots+x_{n}}$
$=\frac{1}{1+x_{1}+\cdots+x_{n}}=\frac{1}{1+a}$.
Thus, $c \geqslant 1-\left(\frac{1}{1+a}\right)^{\frac{1}{n}}$.
Therefore, the minimum value of $c$ is $1-(1+a)^{-\frac{1}{n}}$.
For this minimum value, it can be achieved if and only if the $n$ inequalities in (1) all hold with equality, which means
$$
\frac{1+x_{1}+\cdots+x_{j-1}}{1+x_{1}+\cdots+x_{j}}=(1+a)^{-\frac{1}{n}}, 1 \leqslant j \leqslant n .
$$

Let $S_{1}=1+x_{1}+\cdots+x_{j}$, then $S_{j}=(1+a)^{\frac{1}{n}} S_{j-1}$.
Thus, $\left\{S_{1}\right\}$ is a geometric sequence, and at this point, $S_{1}=(1+a)^{\frac{1}{n}} S_{0}$ (where $S_{0}=1$). Therefore,
$$
x_{j}=S_{j}-S_{j-1}=(1+a)^{\frac{j}{n}}-(1+a)^{\frac{-1}{n}}, 1 \leqslant j \leqslant n .
$$

Therefore, when $c$ takes the minimum value $1-(1+a)^{-\frac{1}{n}}$,
$$
x_{j}=(1+a)^{\frac{i}{n}}-(1+a)^{\frac{i-1}{n}} .
$$