The polynomial $f(x)=1-x+x^2-x^3+\cdots-x^{19}+x^{20}$ is written into the form $g(y)=a_0+a_1y+a_2y^2+\cdots+a_{20}y^{20}$, where $y=x-4$, then $a_0+a_1+\cdots+a_{20}=$________.
1. **Express the polynomial \( f(x) \) as a geometric series:**
   The given polynomial is:
   \[
   f(x) = 1 - x + x^2 - x^3 + \cdots - x^{19} + x^{20}
   \]
   This is a geometric series with the first term \( a = 1 \) and common ratio \( r = -x \).

2. **Sum the geometric series:**
   The sum of a geometric series \( a + ar + ar^2 + \cdots + ar^n \) is given by:
   \[
   S = \frac{a(1 - r^{n+1})}{1 - r}
   \]
   Here, \( a = 1 \), \( r = -x \), and \( n = 20 \). Therefore, the sum is:
   \[
   f(x) = \frac{1 - (-x)^{21}}{1 - (-x)} = \frac{1 - (-x)^{21}}{1 + x}
   \]

3. **Evaluate \( f(5) \):**
   To find \( a_0 + a_1 + \cdots + a_{20} \), we need to evaluate \( f(5) \):
   \[
   f(5) = \frac{1 - (-5)^{21}}{1 + 5} = \frac{1 - (-5)^{21}}{6}
   \]

4. **Simplify the expression:**
   Since \( (-5)^{21} \) is a large negative number, we have:
   \[
   (-5)^{21} = -5^{21}
   \]
   Therefore:
   \[
   f(5) = \frac{1 - (-5^{21})}{6} = \frac{1 + 5^{21}}{6}
   \]

5. **Conclusion:**
   The value of \( a_0 + a_1 + \cdots + a_{20} \) is \( f(5) \):
   \[
   a_0 + a_1 + \cdots + a_{20} = \frac{1 + 5^{21}}{6}
   \]

The final answer is \(\boxed{\frac{1 + 5^{21}}{6}}\).