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                ### 算法分析 **第一次嘗試的算法核心部分** ~~~ for a in range(0, 1001): for b in range(0, 1001): for c in range(0, 1001): if a**2 + b**2 == c**2 and a+b+c == 1000: print("a, b, c: %d, %d, %d" % (a, b, c)) ~~~ 時間復雜度: `T(n) = O(n*n*n) = O(n3)` **第二次嘗試的算法核心部分** ~~~ for a in range(0, 1001): for b in range(0, 1001-a): c = 1000 - a - b if a**2 + b**2 == c**2: print("a, b, c: %d, %d, %d" % (a, b, c)) ~~~ 時間復雜度: ~~~ T(n) = O(n*n*(1+1)) = O(n*n) = O(n2) ~~~ 由此可見,我們嘗試的第二種算法要比第一種算法的時間復雜度好多的。
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