Example 1 (1997 National High School Competition Question) Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms is no less than 3, and the sum of all terms is $97^{2}$. Then the number of such sequences is ( ).
A. 2
B. 3
C. 4
D. 5
Solution: Choose C. Reason: Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. From the given, we have $n a+\frac{1}{2} n(n-1) d=97^{2}$, which is $n[2 a+(n-1) d]=2 \times 97^{2}$.
Since $n \geqslant 3$, $n$ can only be $97, 2 \times 97, 97^{2}, 2 \times 97^{2}$. Also, because $a \geqslant 0, d \geqslant 0$, we have $2 \times 97^{2}=2 n a+n(n-1) d \geqslant n(n-1) d$.
If $d>0$, then by $d$ being an integer, we get $d \geqslant 1$, which means $2 \times 97^{2} \geqslant n(n-1)$, at this time $n$ can only be 97. Therefore, there are only the following two possibilities:
$$
\begin{array}{l}
\text { (1) } n=97, d=1, a=49 \text {; } \\
\text { (2) } n=97, d=2, a=1 .
\end{array}
$$

If $d=0$, then $n a=97^{2}$, at this time there are also only the following two possibilities:
(3) $n=97, d=0, a=97$;
(4) $n=97^{2}, d=0, a=1$.

Therefore, there are a total of 4 arithmetic sequences that meet the conditions.