For each positive integer $n$, define the function
$$
f(n)=\left\{\begin{array}{cl}
0, & \text { when } n \text { is a perfect square; } \\
{\left[\frac{1}{\{\sqrt{n}\}}\right],} & \text { when } n \text { is not a perfect square. }
\end{array}\right.
$$

where $[x]$ denotes the greatest integer not exceeding $x$, and $\{x\}=x-[x]$. Find: $\sum_{k=1}^{200} f(k)$.

Solution: If $k$ is not a perfect square, then there exists $a \in \mathbf{N}$, such that $a^{2}<k<(a+1)^{2}$, then $a<\sqrt{k}<a+1$, i.e.,
$$
\{\sqrt{k}\}=\sqrt{k}-a \text {. }
$$

Therefore, $\left[\frac{1}{\{\sqrt{k}\}}\right]=\left[\frac{1}{\sqrt{k}-a}\right]=\left[\frac{\sqrt{k}+a}{k-a^{2}}\right]$.
$$
\begin{aligned}
\text { Hence } & \sum_{k=1}^{2+0} f(k) \\
= & 15 \times 0+\sum_{k=2}^{3}\left[\frac{\sqrt{k}+1}{k-1}\right]+\sum_{k=5}^{8}\left[\frac{\sqrt{k}+2}{k-4}\right]+ \\
& \sum_{k=10}^{15}\left[\frac{\sqrt{k}+3}{k-9}\right]+\sum_{k=17}^{24}\left[\frac{\sqrt{k}+4}{k-16}\right]+\cdots+ \\
& \sum_{k=197}^{24}\left[\frac{\sqrt{k}+14}{k-196}\right]+\sum_{k=20}^{2+0}\left[\frac{\sqrt{k}+15}{k-225}\right] \\
= & 0+(2+1)+(4+2+1+1)+ \\
& (6+3+2+1 \times 3)+(8+4+2+2+1 \times 4)+ \\
& (10+5+3+2+2+1 \times 5)+ \\
& (12+6+4+3+2+2+1 \times 6)+ \\
& (14+7+4+3+2+2+2+1 \times 7)+ \\
& (16+8+5+4+3+2+2+2+1 \times 8)+ \\
& (18+9+6+4+3+3+2+2+2+1 \times 9)+ \\
& (20+10+6+5+4+3+2+2+2+2+1 \times 10)+ \\
& (22+11+7+5+4+3+3+2+2+2+ \\
& 2+1 \times 11)+(24+12+8+6+4+4+ \\
& 3+3+2+2+2+2+1 \times 12)+(26+13+8+6+5+ \\
& 4+3+3+2+2+2+2+2+1 \times 13)+ \\
& (28+14+9+7+5+4+4+3+3+2+ \\
& 2+2+2+2+1 \times 14)+(30+15+10+7+ \\
& 6+5+4+3+3+3+2+2+2+2+2) \\
= & 768
\end{aligned}
$$