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第一次小考解答

第一次小考解答. 張啟中. 1. Compute the Address of Array. 由下圖可知，對於陣列元素 A[i][j][k] 而言，其位址為 α+ (i-1) *200 + (j-1) *20 + (k-1) 所以 A[3][7][2] = α+400+120+1 = α+521. A[1][1][1]. A[1][1][2]. A[1][1][3]. ‥‥. A[1][1][19]. A[1][1][20]. α. α+1. α+2. α+18. α+19. A[1][2][1]. A[1][2][2].

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第一次小考解答

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  1. 第一次小考解答 張啟中

  2. 1. Compute the Address of Array • 由下圖可知，對於陣列元素 A[i][j][k] 而言，其位址為 α+(i-1)*200 + (j-1)*20 + (k-1) • 所以 A[3][7][2] = α+400+120+1 = α+521 A[1][1][1] A[1][1][2] A[1][1][3] ‥‥ A[1][1][19] A[1][1][20] α α+1 α+2 α+18 α+19 A[1][2][1] A[1][2][2] A[1][2][3] ‥‥ A[1][2][19] A[1][2][20] α+20 α+21 α+22 α+38 α+39 A[1][10][1] A[1][10][2] A[1][10][3] ‥‥ A[1][10][19] A[1][10][20] α+180 α+181 α+182 α+198 α+199 A[2][1][1] A[2][1][2] A[2][1][3] ‥‥ A[2][1][19] A[2][1][20] α+200 α+201 α+202 α+218 α+219

  3. 3+1=4 0+1=1 -1+1=0 2. Compute the Failure Function b ≠c a ≠c 0 1 2 3 4 5 6 7 8 9 a b c a b c a c a b 3 -1 -1 -1 0 1 2 -1 ?

  4. 2. Compute the Failure Function 0 1 2 3 4 5 6 7 8 9 a b c a b c a c a b -1 -1 -1 0 1 2 3 ?

  5. 2. Compute the Failure Function 0 1 2 3 4 5 a a a a a b -1 -1 0 1 2 3 ?

  6. 2. Compute the Failure Function void String::fail() //Compute the failure function for the pattern p(*this) //p is a string { intLengthP = Length(); f[0] = -1; for(int j=1; j< LengthP; j++) //compute f[j] { int i=f[j-1]; while (p[j] != p[i+1] && i>=0) i = f[i]; if (p[j] == p[i+1]) f[j] = i+1; else f[j] = -1; } } //end of fail O(LengthP)

  7. -1 -1 -1 0 1 2 3 -1 0 1 a b c a b c a c a b a b c a b c a b c a b c a c a b 2. String Pattern Matching 3+1=4 b ≠ c

  8. -1 -1 -1 0 1 2 3 -1 0 1 a b c a b c a c a b a b c a b c a b c a b c a c a b 2. String Pattern Matching 3+1=4 OK b ≠ c 懂了沒？

  9. -1 -1 -1 0 1 2 3 -1 0 1 a b c a b c a c a b a b c a b c a a c a b c a c a b 2. String Pattern Matching 3+1=4 a ≠ c

  10. -1 -1 -1 0 1 2 3 -1 0 1 a b c a b c a c a b a b c a b c a a c a b c a c a b 2. String Pattern Matching 0+1=1 -1+1=0 a ≠ b

  11. -1 -1 -1 0 1 2 3 -1 0 1 a b c a b c a c a b a b c a b c a a c a b c a c a b 2. String Pattern Matching -1+1=0 剩下的自己追蹤了!

  12. 2. String Pattern Matching int String::FastFind(String pat) { //Determine if pat is a substring of s int PosP=0, PosS=0; int LengthP=pat.length(), LengthS=length(); while ((PosP<LengthP)&&(PosS<LengthS)) if (pat[PosP] == str[PosS]) PosP++; PosS++; else if (PosP == 0) PosS++; else PosP = pat.f[PosP-1]+1; if (PosP < LengthP) return -1; else return PosS-LengthP; } //end of FastFind O(LengthS)

  13. 1 2 4 8 16 32 0 1 2 3 4 5 0 2 8 24 64 160 1 4 16 64 256 1024 1 8 64 512 4096 32,768 2 4 16 256 65,536 4,294,967,296 1 2 24 40,320 20,922,789,888,000 ……. 3. Order the Time Complexity O(1) < O(logn) < O(n) < O(nlogn) < O(n2) < O(n3) < O(2n) < O(n!)

  14. 4. Show that 10n2+9 = O(n2) • By Big-O definition f (n) = O(g(n)) 若且唯若存在有 c＞0和 n0N, 對所有的 n, n≧n0 使得 f (n) ≦ c．g(n) 。 • 本題只要找出 c與 n0即可得證 取 c = 11且 n0 = 3，則對所有的 n, n≧3會使得 10n2 + 9 ≦ 11．g(n) = 11n2 即 10n2+9 = O(n2) 得證

  15. 5. Recursive Program n! int factor(int n) { if (n==0 || n==1) return 1; else return n*factor(n-1); }

  16. 5. Recursive Program Permutation //pos 表示目前要排列位置，從 pos(含)以後的字元需要被排列 void permutation(string s, int pos) { if (pos == s.length()-1) { cout << s << endl; } else { for (int i=pos; i<s.length(); i++) { swap(s[i],s[pos]); //交換 s[i] 與 s[pos] permutation(s, pos+1); swap(s[i],s[pos]); //交換 s[i] 與 s[pos] } } } 想想看，本程式的時間複雜度為多少？

  17. 6. Compute the Recursive Formula • T(n) = 2T(n-1) + 1 and T(1)=1 T(n) = 2T(n-1)+1 = 2[ 2T(n-2)+1 ] + 1 = 22 T(n-2) + 2 +1 = 22[ 2T(n-3)+1 ] + 2 + 1 = 23T(n-3) + 22+ 2+ 1 = …… = 2n-1 T(1) + 2n-2 + 2n-3 + …. + 2 + 1 = 2n-1 + 2n-2 + …. + 2 + 1  (等比級數，公比=2，有n 項) = 2n – 1

  18. 7. 表示式的運算 • 注意這三小題的區別。 • 人工筆算 將中序表示式 (A+B)*D + E/(F+A*D)+C轉換為後序表示式 • 電腦運算 利用堆疊計算後序表示式，得到結果。 • 電腦運算 利用堆疊將一個中序表示式轉換為後序表示式。 請注意第 2 與 3 小題的區別

  19. 7. 表示式的運算 (A+B)*D + E/(F+A*D)+C • AB+*D+E/(F+AD*)+C • AB+D*+E/FAD*++C • AB+D*+EFAD*+/+C • AB+D*EFAD*+/++C • AB+D*EFAD*+/+C+

  20. 2 3 4 + * 4 + * * 7 2 3 2 3 4 + * + * 4 3 2 2 14 7. 表示式的運算

  21. 7. 表示式的運算 中置表示式轉後置表示式 1. 當碰到運算元時，直接輸出此運算元。 2. 當碰到左括號時，將其 push 至堆疊中。 3. 當碰到右括號時，將堆疊中的運算子 pop 出來並輸出，直到碰到左括號為止，然後再將此左括號 pop 掉。 4. 當碰到運算子時， 依序將堆疊中運算優先次序較高或相同的運算子 pop 出來並輸出，直到遇到下列情形之一 (1) 碰到左括號 (2) 碰到優先次序較低的運算子 (3) 堆疊變為空 最後將此運算子 push 至堆疊中。 5. 當輸入字串全部處理完時，將堆疊中所剩餘的運算子逐一地 pop 出來並輸出，直到堆疊變為空為止。

  22. 0 1 2 3 4 5 0 1 2 3 4 5 8. Sparse Matrix

  23. 8. Sparse Matrix SparseMatrix SparseMatrix::Transpose() // The transpose of a(*this) is placed in b and is found // in O(terms + columns) time. { int *Rows = newint[Cols]; int *RowStart = newint[Rows]; SparseMatrix b; b.Rows = Cols; b.Cols = Rows; b.Terms = Terms; if (Terms > 0) // nonzero matrix { //compute RowSize[i] = number of terms in row i of b // initialize for (int i = 0; i < Cols; i++) RowSize[i] = 0; for (int i = 0; i < Terms; i++) RowSize[smArray[i].col]++; // RowStart[i] = starting position of row i in b RowStart[0] = 0; for (int i = 1; i < Cols; i++) RowStart[i] = RowStart[i-1] + RowSize[i-1]; ……… } } // end of FastTranspose

  24. 8. Sparse Matrix

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