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Problem #1
Expression Value
-----------------------------------------------
3 * 4 + 5 * 6 42
23 % 5 - 17 % (16 % 10) -2
"1" + 2 + 3 * 4 + (5 + 6) "121211"
1.5 * 2 + 20 / 3 / 4.0 + 6 / 4 5.5
345 / 10 / 3 + 10 / (5 / 2.0) 15.0
2 points each, 1 point for correct value of wrong type (missing parens, no .0)
-------------------------------------------------------------------------------
Problem #2:
solution:
foo wants buzz to sue.
bill wants sue to buzz.
bill wants sue to hope.
hope wants bill to sue.
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(1, 8); 3 8
ifElseMystery(3, 5); 5 0
ifElseMystery(4, 5); 5 6
ifElseMystery(8, 6); 8 2
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 Value Returned
--------------------------------------
mystery(3) 2
mystery(4) 3
mystery(5) 5
mystery(6) 8
3 points each
Problem #5:
y > x z < 0 z > 0
+---------------------+---------------------+---------------------+
Point A | sometimes | never | never |
+---------------------+---------------------+---------------------+
Point B | never | never | sometimes |
+---------------------+---------------------+---------------------+
Point C | sometimes | never | always |
+---------------------+---------------------+---------------------+
Point D | always | never | sometimes |
+---------------------+---------------------+---------------------+
Point E | always | sometimes | sometimes |
+---------------------+---------------------+---------------------+
1 point per blank
Problem #6
Solution:
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();
}
2 header (-1 per error)
2 printing leading text ("The first...");
1 reasonable attempt (includes both parameters and some of the text)
1 correct (all correct, not inside or after the loop)
1 exactly one println at the end
3 looping m times (either loop goes m times or fencepost + loop goes m-1 times)
1 any kind of loop
1 within obob of correct bounds
1 correct bounds
3 printing the multiples of n inside a loop
1 reasonable attempt (prints something that involves n and i)
2 correct (correctly computing and printing the multiples with commas)
4 solved the fencepost
1 attempt (printed one of the numbers outside the loop, this would have
affected their loop bounds scoring)
3 correct (requires printing commas in a loop and loop bounds are within
obob; mostly atomic, -1 if minor error, ask question leader about -1)
Problem #7
Solution:
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;
}
1 correct header
4 loop to get digits
1 attempt to use % 10 to get a digit
1 correctly uses % 10 to get a digit
1 correctly updates n to n / 10
1 correct test (n != 0 or n > 0)
8 max/min computation
1 some attempt at max/min computation
3 correct initialization of max and min
1 max initialized to a digit or something <= 0
1 min initialized to a digit or something >= 9
1 all correct: either a digit or works in all cases including 0
2 correctly updates min (min digit) in the loop
2 correctly updates max (max digit) in the loop
2 returning their result
1 returns something
1 correctly returns (max digit) - (min digit), but no double jeopardy if
they had an error in computing those OR returns a max of some digit delta
(e.g., difference of adjacent digits)
Problem #8
Solutions:
public static String undouble(String s) {
String result = "";
if (s.length() > 0) {
char last = s.charAt(0);
result += last;
for (int i = 1; i < s.length(); i++) {
if (s.charAt(i) != last) {
result += s.charAt(i);
}
last = s.charAt(i);
}
}
return result;
}
public static String undouble(String s) {
String result = "";
if (s.length() > 0) {
result = s.substring(0, 1);
for (int i = 1; i < s.length(); i++) {
if (s.charAt(i) != s.charAt(i - 1)) {
result += s.charAt(i);
}
}
}
return result;
}
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 str += charAt or substring
1 comparing charAt(i) with charAt(i + 1) (or substring)
If both of the last two (concat & adjacent comparison), then give to Stuart to
grade for score of 4-9.
-1 if fails on empty string or case like ccaatt
-2 if their code doesn't handle 1st/last char properly
-0 String result = s.charAt(0);
-0 for String new = ... (reserved word)
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Stuart Reges
Last modified: Fri Sep 16 16:16:07 PDT 2011