fix R8a and add R8*.md and build them

reviews/R8c.md

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+# Review 8-3
+
+* Hajin Ju, 2024062806
+
+## Problem 1
+
+Fill in the blanks in the following LCS computation
+
+### Solution 1
+
+$$c[i][j] = \text{The length of an LCS of the subsequences} \,X_i\, \text{and}\, Y_j.$$
+
+$$c[i][j] = \begin{cases}
+0 & \text{if}\, i = 0 \,\text{or}\, j = 0\\
+c[i-1][j-1] + 1 &\text{if}\, i,j > 0 \,\text{and}\, x_i = y_j\\
+max(c[i][j-1], c[i-1][j]) &\text{if}\, i, j > 0 \,\text{and}\, x_i \neq y_j
+\end{cases}$$
+
+| *     | $y_j$ | $B$          | $D$           | $C$           | $A$            | $B$            | $A$            |
+| ----- | ----- | ------------ | ------------- | ------------- | -------------- | -------------- | -------------- |
+| $x_i$ | 0     | 0            | 0             | 0             | 0              | 0              | 0              |
+| $A$   | 0     | $\uparrow 0$ | $\uparrow 0$  | $\uparrow 0$  | $\nwarrow 1$   | $\leftarrow 1$ | $\nwarrow 1$   |
+| $B$   | 0     | $\nwarrow1$  | $\leftarrow1$ | $\leftarrow1$ | $\uparrow 1$   | $\nwarrow 2$   | $\leftarrow 2$ |
+| $C$   | 0     | $\uparrow 1$ | $\uparrow 1$  | $\nwarrow 2$  | $\leftarrow 2$ | $\uparrow 2$   | $\uparrow 2$   |
+| $B$   | 0     | $\nwarrow 1$ | $\uparrow 1$  | $\uparrow 2$  | $\uparrow2$    | $\nwarrow3$    | $\leftarrow3$  |
+| $D$   | 0     | $\uparrow1$  | $\nwarrow2$   | $\uparrow2$   | $\uparrow2$    | $\uparrow3$    | $\uparrow3$    |
+| $A$   | 0     | $\uparrow1$  | $\uparrow2$   | $\uparrow2$   | $\nwarrow 3$   | $\uparrow 3$   | $\nwarrow4$    |
+| $B$   | 0     | $\nwarrow1$  | $\uparrow2$   | $\uparrow2$   | $\uparrow3$    | $\nwarrow4$    | $\uparrow4$    |
+
+## Problem 2
+
+Fill in the blanks in the following multiple LCS computation.
+
+### Solution 2
+
+| *     | $y_j$ | $B$                    | $D$                    | $C$                    | $A$                    | $B$                    | $A$                    |
+| ----- | ----- | ---------------------- | ---------------------- | ---------------------- | ---------------------- | ---------------------- | ---------------------- |
+| $x_i$ | 0     | 0                      | 0                      | 0                      | 0                      | 0                      | 0                      |
+| $A$   | 0     | $\leftarrow\uparrow 0$ | $\leftarrow\uparrow 0$ | $\leftarrow\uparrow 0$ | $\nwarrow 1$           | $\leftarrow 1$         | $\nwarrow 1$           |
+| $B$   | 0     | $\nwarrow1$            | $\leftarrow1$          | $\leftarrow1$          | $\leftarrow\uparrow 1$ | $\nwarrow 2$           | $\leftarrow 2$         |
+| $C$   | 0     | $\uparrow 1$           | $\leftarrow\uparrow 1$ | $\nwarrow 2$           | $\leftarrow 2$         | $\leftarrow\uparrow 2$ | $\leftarrow\uparrow 2$ |
+| $B$   | 0     | $\nwarrow 1$           | $\leftarrow\uparrow 1$ | $\uparrow 2$           | $\uparrow2$            | $\nwarrow3$            | $\leftarrow3$          |
+| $D$   | 0     | $\uparrow1$            | $\nwarrow2$            | $\leftarrow\uparrow2$  | $\uparrow2$            | $\uparrow3$            | $\uparrow3$            |
+| $A$   | 0     | $\uparrow1$            | $\uparrow2$            | $\uparrow2$            | $\nwarrow 3$           | $\leftarrow\uparrow 3$ | $\nwarrow4$            |
+| $B$   | 0     | $\nwarrow1$            | $\uparrow2$            | $\leftarrow\uparrow2$  | $\uparrow3$            | $\nwarrow4$            | $\leftarrow\uparrow4$  |
+
+
+## Problem 3
+
+Fill in the blanks in the following pseudocode for `LCS-LENGTH`.
+
+### Solution 3
+
+```text
+LCS-LENGTH(X, Y)
+    m = X.length
+    n = Y.length
+    let b[1..m, 1..n] and c[1..m, 1..n] be new tables
+    for i = 1 to m
+        c[i][0] = 0
+    for j = 1 to n
+        c[0][j] = 0
+    for i = 1 to m
+        for j = 1 to n
+            if X[i] = Y[j]
+                c[i][j] = c[i - 1][j - 1] + 1
+                b[i][j] = \nwarrow
+            else if c[i-1][j] >= c[i][j-1]
+                c[i][j] = c[i-1][j]
+                b[i][j] = \uparrow
+            else
+                c[i][j] = c[i][j-1]
+                b[i][j] = \leftarrow
+    return c, b
+```
+
+## Problem 4
+
+Fill in the blanks in the following pseudocode for `PRINT-LCS`.
+
+### Solution 4
+
+```text
+PRINT-LCS(b, X, i, j)
+    if i = 0 or j = 0
+        return
+    if b[i][j] == \nwarrow
+        PRINT-LCS(b, X, i-1, j)
+        print X[i]
+    else if b[i][j] == \uparrow
+        PRINT-LCS(b, X, i-1, j)
+    else
+        PRINT-LCS(b, X, i, j-1)
+```