function f(n)
if n == 0
return 0
else if n == 1
return 1
else
return f(n-1)+f(n-2)how about f(1000000)???
too slow!!! TLE
但是這不是今天的重點
把算過的東西記起來
重點
不要重複計算
f[0] = 0
f[1] = 1
for i = 2~N
f[i] = f[i-1] + f[i-2]only n-1 times addition for f(n)!! fast
faster?
yes!! fast exponential