Xiaopang arranged the 50 integers $1, 2, 3, 4, ..., 49, 50$ in ascending order without any gaps, and then inserted a “+” between each pair of adjacent numbers, resulting in an addition expression: $1+2+3+4+5+6+7+8+9+1+0+1+1+\cdots+4+9+5+0$. Please calculate the sum of this addition expression, the result is $\qquad$ _
【Analysis and Solution】This problem is to find the sum of the digits of the integers from 1 to 50.
(Method One)
In the units place, the digits $1, 2, \ldots, 9, 0$ each appear 5 times;
In the tens place, the digits $1, 2, 3, 4$ each appear 10 times, and 5 appears only once;
Thus, the sum of the digits of the integers from 1 to 50 is $(1+2+\ldots+9+0) \times 5+[(1+2+3+4) \times 10+5 \times 1]=330$;
Therefore, the result is 330.
(Method Two)
We first find the sum of the digits of the integers from 0 to 49:
Since $0+49, 1+48, \ldots, 24+25$ do not produce any carry;
Thus, the sum of the digits of the integers from 0 to 49 is $(4+9) \times (50 \div 2)=325$;
Therefore, the sum of the digits of the integers from 1 to 50 is $325-0+(5+0)=330$;
Therefore, the result is 330.