QuestionA
宣告出一個vector,讓他去儲存n頭牛的重量, 再利用sort()函式將接收到的數字由小排到大, 之後用一個for迴圈,從第n-1格(vector)一直加到第n-6格 如此把前五重的牛重量相加,最後輸出加完的結果
QuestionB
1000 10000 100000 1000000
(nlogn) sort() (seconds): 0.000164 0.00216 0.026709 0.303657
(n^2) insert()(seconds): 0.007999 0.768312 76.6689 7593.8
依據結果發現,執行sort()所需的時間小於insert function所需的時間, 且當所需排序的個數越大時,兩者的時間差也越大, 明顯看出O(nlogn)及O(n^2)所需時間的差異
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