One, (20 points) A TV station, to understand the viewing situation of its two characteristic programs $A, B, C$, surveyed 28 viewers and found: each viewer watched at least one of the programs; among the viewers who did not watch program $A$, the number of viewers who watched program $B$ was twice the number of viewers who watched program $C$; among the viewers who watched program $A$, the number of viewers who only watched program $A$ was one more than the number of viewers who watched program $A$ and at least one other program; among the viewers who only watched one program, half did not watch program $B$ or $C$. Find the viewership rate of program $A$.
Let the number of viewers who only watch programs $A, B, C$ be $x, y, z$ respectively, the number of viewers who do not watch program $A$ but watch programs $B$ and $C$ be $m$, and the number of viewers who do not only watch program $A$ be $n$, as shown in Figure 7.
From the problem, we have
$$
\left\{\begin{array}{l}
y+m=2(z+m), \\
n=x-1, \\
x=y+z, \\
x+n+y+m+z=28 .
\end{array}\right.
$$

From equation (1), we get
$$
m=y-2 z \text {. }
$$

Substituting equations (2), (3), and (5) into equation (4), we get
$$
\begin{array}{l}
(y+z)+(y+z-1)+y+(y-2 z)+z \\
=28,
\end{array}
$$

which simplifies to $z=29-4 y$.
From equation (5), we know $y-2 z=m \geqslant 0$, which implies $9 y-58 \geqslant 0 \Rightarrow y \geqslant \frac{58}{9}$.
From equation (6), we know
$29-4 y=z \geqslant 0$
$\Rightarrow 4 y \leqslant 29 \Rightarrow y \leqslant \frac{29}{4}$.
Combining these, and considering $y$ is an integer, we get $y=7$.
Thus, $z=1, x=8, n=7, m=5$.
Therefore, the viewership rate of program $A$ is
$$
\frac{x+n}{28} \times 100 \%=\frac{15}{28} \times 100 \%=53.6 \% \text {. }
$$