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Problem #1
Expression Value
-------------------------------------------------
16 / 3 + 3.2 * 2 11.4
8 / 7 + "5 / 4" + 8 / 3 "15 / 42"
88 % 10 % 3 * 16 / 10 3
29 / 3 / 2 / 4.0 + 3.6 * 2 8.2
1.4 + (3 + 2 * 6) / (8 - 14 / 3) * 2.2 8.0
2 points each, 1 point for correct value of wrong type (missing parens, no .0)
-------------------------------------------------------------------------------
Problem #2:
solution:
to walk the walk is good
to hear the good is bad
to feel the walk is song
to feel the talk is bad
1 point for each of the 12 values
-1 if any error in surrounding text (dittos not allowed)
-4 if two columns transposed
-8 if you need to transpose twice
-0 if quoted
-------------------------------------------------------------------------------
Problem #3:
Method Call Output Produced
---------------------------------------
ifElseMystery(4, 15); 6 15
ifElseMystery(7, 17); 9 17
ifElseMystery(12, 5); 12 1
ifElseMystery(16, 8); 16 4
each is 3 points, -1 for a single number wrong or for a single transpose
-2 overall if columns transposed
-2 if formatting error (e.g., commas between numbers)
-1 if individual pair is reversed
-0 if output is quoted
-0 if underscore for space
-------------------------------------------------------------------------------
Problem #4:
Method Call Output Produced
---------------------------------------
mystery(4); 2 2
mystery(5); 1 5
mystery(24); 4 3
mystery(28); 3 7
column 1 is 1 point, column 2 is 2 points
-2 overall if columns transposed
-2 if formatting error (e.g., commas between numbers)
-1 if individual pair is reversed
-0 if output is quoted
-0 if underscore for space
Problem #5:
x > y z == 0 x == y
+---------------------+---------------------+---------------------+
Point A | sometimes | always | sometimes |
+---------------------+---------------------+---------------------+
Point B | sometimes | sometimes | never |
+---------------------+---------------------+---------------------+
Point C | always | never | never |
+---------------------+---------------------+---------------------+
Point D | never | never | never |
+---------------------+---------------------+---------------------+
Point E | never | sometimes | always |
+---------------------+---------------------+---------------------+
1 point per blank
Problem #6
Solution:
public static void printTwoDigit(Random r, int n) {
boolean seen42 = false;
for (int i = 1; i <= n; i++) {
int number = r.nextInt(90) + 10;
System.out.println("next = " + number);
if (number == 42) {
seen42 = true;
}
}
if (seen42) {
System.out.println("We saw a 42");
} else {
System.out.println("no 42");
}
}
2 header (-1 per error)
2 looping n times
1 within obob of correct bounds
1 correct bounds
4 choosing random 2-digit number
1 reasonable attempt to generate a 2-digit number somewhere
1 correct expression for generating 2-digit number somewhere
2 computes a new random value each time through the loop
2 println inside the loop to print 2-digit number (no double jeopardy with
choosing the numbers to print)
1 attempt
1 correct
5 Did we see a 42?
1 they have both printlns ("We saw a 42" and "no 42")
2 some kind of flag or counter variable that is updated when 42 is seen
(exception: -1 for small mistakes, such as checking for 24 instead of 42)
1 correct initialization of that counter/flag
1 counter/flag variable used correctly with an if/else or if/if after the
loop to choose between the two printlns
Problem #7
Solutions:
public static boolean hasOddEvenDigit(int n) {
boolean hasEven = false;
boolean hasOdd = false;
while (n > 0) {
int digit = n % 10;
n = n / 10;
if (digit % 2 == 0) {
hasEven = true;
} else {
hasOdd = true;
}
}
return hasEven && hasOdd;
}
public static boolean hasOddEvenDigit2(int n) {
int evens = 0;
int odds = 0;
while (n > 0) {
if (n % 2 == 0) {
evens++;
} else {
odds++;
}
n = n / 10;
}
return evens > 0 && odds > 0;
}
1 correct header
1 uses % 10 to get the last digit or % 2 to find out if last digit is even
4 loop to examine digits (don't worry about potential early return for this)
2 correctly updates n to n / 10 once each time through loop
2 correct test to process all digits: n != 0 or n > 0
(-1 if obob like n / 10 != 0 or bad placement of n = n / 10)
6 recognizing odd/even digits
3 state variable to keep track of whether an even digit has been seen
1 declared and initialized properly
2 updated properly inside the loop
3 state variable to keep track of whether an odd digit has been seen
1 declared and initialized properly
2 updated properly inside the loop
note: if the variables are never declared/initialized or declared only
inside the if, then they lose both of the declare/initialize points AND a
syntax point, but they can get the update points
3 returning their result
1 reasonable attempt to return an appropriate value (can include checking
adjacent digits for an odd/even pair or returning after a loop that is
testing for odd/even digits)
2 correctly uses odd/even state variables to return whether both were seen
(lose these points if they return early, -1 if minor error--ask question
leader if it's a minor error)
Problem #8
Solutions:
public static void printStripped(String s) {
boolean inComment = false;
for (int i = 0; i < s.length(); i++) {
char next = s.charAt(i);
if (next == '<') {
inComment = true;
} else if (inComment && next == '>') {
inComment = false;
} else if (!inComment) {
System.out.print(next);
}
}
System.out.println();
}
public static void printStripped(String s) {
int start = s.indexOf('<');
while (start != -1) {
int stop = s.indexOf('>', start + 1);
s = s.substring(0, start) + s.substring(stop + 1);
start = s.indexOf('<');
}
System.out.println(s);
}
Can get up to 3 points for any of the following good things:
1 correct header
1 for loop involving s.length
1 returning a String
1 s.charAt(i) or s.substring(i, i + 1)
1 s.indexOf("<") or s.indexOf(">")
1 printing an individual character
1 printing a substring that involved s.indexOf("<") or s.indexOf(">")
1 some kind of flag to keep track of whether you are in a comment
To make it to round 2, they have to have either some kind of flag that keeps
track of whether or not they are in a comment (comparing against "<" and ">")
or combinations of s.subtring and s.indexOf("<") and s.indexOf(">")
]]>
Stuart Reges
Last modified: Fri Sep 16 16:16:07 PDT 2011