$10 \cdot 54$ There are 1000 numbers arranged in a row, where among any three consecutive numbers, the middle number is equal to the sum of the numbers before and after it. If the first number and the second number are both 1, then the sum of these 1000 numbers is
(A) 1000.
(B) 1.
(C) 0.
(D) -1.
(China Shandong Province Junior High School Mathematics Competition, 1996)
［Solution］According to the problem, we know that the 1000 numbers are arranged in the following manner:
$1,1,0,-1,-1,0,1,1,0,-1,-1,0, \cdots$
In other words, they cycle periodically with the sequence $1,1,0,-1,-1,0$.
Also, $1000 \div 6=166 \cdots \cdots 4$, and the sum of the 6 numbers in each cycle is 0, so the sum of these 1000 numbers is
$$
1+1+0+(-1)=1 \text {. }
$$

Therefore, the answer is $(B)$.