给出序列an=an-1+an-2+an-3,n=3,4,5…,a0=1,a1=2,a2=3的递归求解算法解:
procedure sequence(n : nonnegative integer)
if n < 3 then return n + 1
else return sequence(n - 1) + sequence(n - 2) + sequence(n - 3)
给出序列an=an-1+an-2+an-3,n=3,4,5…,a0=1,a1=2,a2=3的递归求解算法解:
procedure sequence(n : nonnegative integer)
if n < 3 then return n + 1
else return sequence(n - 1) + sequence(n - 2) + sequence(n - 3)
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