We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
1. 下列說法正確的是 （ ）
A. Java語言不區分大小寫
B. Java程序以類為基本單位
C. JVM為Java虛擬機JVM的英文縮寫
D. 運行Java程序需要先安裝JDK
2. 下列說法中錯誤的是 ( )
A. Java語言是編譯執行的
B. Java中使用了多進程技術
C. Java的單行注視以//開頭
D. Java語言具有很高的安全性
3. 下面不屬于Java語言特點的一項是( )
A. 安全性
B. 分布式
C. 移植性
D. 編譯執行
4. 下列語句中，正確的項是 ( )
A . int $e,a,b=10
B. char c,d=’a’
C. float e=0.0d
D. double c=0.0f
Instead of finding the longest common
subsequence, let us try to determine the
length of the LCS.
Then tracking back to find the LCS.
Consider a1a2…am and b1b2…bn.
Case 1: am=bn. The LCS must contain am,
we have to find the LCS of a1a2…am-1 and
b1b2…bn-1.
Case 2: am≠bn. Wehave to find the LCS of
a1a2…am-1 and b1b2…bn, and a1a2…am and
b b b
b1b2…bn-1
Let A = a1 a2 … am and B = b1 b2 … bn
Let Li j denote the length of the longest i,g g
common subsequence of a1 a2 … ai and b1 b2
… bj.
Li,j = Li-1,j-1 + 1 if ai=bj
max{ L L } a≠b i-1,j, i,j-1 if ai≠j
L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
特點(FEATURES) 精確度0.1%滿刻度 (Accuracy 0.1%F.S.) 可作各式數學演算式功能如:A+B/A-B/AxB/A/B/A&B(Hi or Lo)/|A| (Math functioA+B/A-B/AxB/A/B/A&B(Hi&Lo)/|A|/etc.....) 16 BIT 類比輸出功能(16 bit DAC isolating analog output function) 輸入/輸出1/輸出2絕緣耐壓2仟伏特/1分鐘(Dielectric strength 2KVac/1min. (input/output1/output2/power)) 寬范圍交直流兩用電源設計(Wide input range for auxiliary power) 尺寸小,穩定性高(Dimension small and High stability)
Description: FASBIR(Filtered Attribute Subspace based Bagging with Injected Randomness) is a variant of Bagging algorithm, whose purpose is to improve accuracy of local learners, such as kNN, through multi-model perturbing ensemble.
Reference: Z.-H. Zhou and Y. Yu. Ensembling local learners through multimodal perturbation. IEEE Transactions on Systems, Man, and Cybernetics - Part B: Cybernetics, 2005, vol.35, no.4, pp.725-735.
Description: S-ISOMAP is a manifold learning algorithm, which is a supervised variant of ISOMAP.
Reference: X. Geng, D.-C. Zhan, and Z.-H. Zhou. Supervised nonlinear dimensionality reduction for visualization and classification. IEEE Transactions on Systems, Man, and Cybernetics - Part B: Cybernetics, 2005, vol.35, no.6, pp.1098-1107.
