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Key to CSE143X Sample Midterm, Fall 2016
1. Expression Value
-----------------------------------------------
5 * 6 - (4 + 3) * 2 - 2 * 3 10
208 / 20 / 4 + 12 / 10.0 + 0.4 * 2 4.0
8 - 2 + "8 - 2" + (8 - 2) + 8 "68 - 268"
4 * 5 % 6 + 297 % 10 + 4 % 8 13
13 / 2 * 3.0 + 5.5 * 3 / 2 26.25
2. Parameter Mystery. The program produces the following output.
he says can we yes
Barry says maybe we can
we says he we can
Sarah says no we can't
3. Method Call Output Produced
---------------------------------------
ifElseMystery(14, 14); 14 13
ifElseMystery(4, 5); 7 5
ifElseMystery(10, 5); 12 10
ifElseMystery(2, 8); 3 8
4. Method Call Output Produced
---------------------------------------
mystery(0); 1 1
mystery(7); 2 12
mystery(32); 3 123
mystery(256); 4 1234
5. x < y z == 0 z % 2 == 0
+---------------------+---------------------+---------------------+
Point A | sometimes | always | always |
+---------------------+---------------------+---------------------+
Point B | always | sometimes | sometimes |
+---------------------+---------------------+---------------------+
Point C | sometimes | never | always |
+---------------------+---------------------+---------------------+
Point D | sometimes | never | never |
+---------------------+---------------------+---------------------+
Point E | never | sometimes | sometimes |
+---------------------+---------------------+---------------------+
6. Programming. One possible solution appears below.
public static int digitRange(int n) {
int min = n % 10;
int max = n % 10;
while (n > 0) {
int digit = n % 10;
n = n / 10;
if (digit < min) {
min = digit;
}
if (digit > max) {
max = digit;
}
}
return max - min;
}
7. File Processing. One possible solution appears below.
public static void printDuplicates(Scanner input) {
while (input.hasNextLine()) {
String line = input.nextLine();
Scanner lineScan = new Scanner(line);
String token = lineScan.next();
int count = 1;
while (lineScan.hasNext()) {
String token2 = lineScan.next();
if (token2.equals(token)) {
count++;
}
if (count > 1 && (!lineScan.hasNext() || !token2.equals(token))) {
System.out.print(token + "*" + count + " ");
count = 1;
}
token = token2;
}
System.out.println();
}
}
8. Arrays. One possible solution appears below.
public static boolean hasAlternatingParity(int[] list) {
for (int i = 0; i < list.length - 1; i++) {
if (list[i] % 2 == list[i + 1] % 2) {
return false;
}
}
return true;
}
9. Programming. Two possible solutions appear below.
public static void distributeTokens(int[] tokens, int player) {
int n = tokens[player];
tokens[player] = 0;
for (int i = 1; i <= n; i++) {
tokens[(player + i) % tokens.length]++;
}
}
public static void distributeTokens(int[] tokens, int player) {
int num = tokens[player];
tokens[player] = 0;
while (num > 0) {
player++;
if (player == tokens.length) {
player = 0;
}
tokens[player]++;
num--;
}
}
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