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Key to CSE143X Midterm, Fall 2013 handout #11
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. One possible solution appears below.
public static void printMultiples(int n, int times) {
System.out.print("The first " + times + " multiples of " + n
+ " are " + + n);
for (int i = 2; i <= times; i++) {
System.out.print(", " + i * n);
}
System.out.println();
}
7. 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;
}
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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