For each positive integer $n$, let $f(n)$ denote the last digit of $1+2+\cdots+n$ (for example, $f(1)=1$, $f(2)=3$, $f(3)=6$). Calculate the value of $f(1)+f(2)+\cdots+f(2011)$.
One, because the last digit of the sum of any 20 consecutive positive integers is 0, so,
$$
f(n+20)=f(n) .
$$

Thus, it is only necessary to calculate the $f(n)$ corresponding to 1 to 20, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$f(n)$ & 1 & 3 & 6 & 0 & 5 & 1 & 8 & 6 & 5 & 5 \\
\hline$n$ & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline$f(n)$ & 6 & 8 & 1 & 5 & 0 & 6 & 3 & 1 & 0 & 0 \\
\hline
\end{tabular}

Also, $f(1)+f(2)+\cdots+f(20)=70$,
$2011=20 \times 100+11$,
then $f(1)+f(2)+\cdots+f(2011)$
$=70 \times 100+f(1)+f(2)+\cdots+f(11)$
$=7000+46=7046$.