1. Write the 4 numbers $1,9,8,8$ in a row and perform the following operation: For each pair of adjacent numbers, subtract the left number from the right number, and write the result between the two numbers. This completes one operation. Then, perform the same operation on the 7 numbers that are now in a row, and continue this process for a total of 100 operations. Find the sum of the final row of numbers.
（Problem from the 51st Moscow Mathematical Olympiad）
1. Each time an operation is performed, the leftmost number is only subtracted once, and the rightmost number is only subtracted from once, while each of the middle numbers is both subtracted and subtracted from once. Therefore, each operation increases the sum of the numbers in the row by $8-1=7$. Hence, after 100 operations, the sum of the numbers in the row is $S=1+9+8+8+7 \times 100=726$.