14S-Section-Solution

14S-Section-Solution - CS106A Handout 14S April 13th 15th,...

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CS106A Handout 14S Spring 2011 April 13 th – 15 th , 2010 Section Solution Discussion Problem 1 Solution: Precedence The answer with Java’s precedence rules is –50. Discussion Problem 2 Solution: Perfect and Near Perfect Numbers Please remember that there is no single approach to solving these problems, and that provided you generate the expected results and the code is easy to understand, it’s all good. public class PerfectAndNearPerfectNumbers extends ConsoleProgram { public void run() { println("This program prints all of the perfect numbers between 1 and 10000"); println(); int numPerfect = 0; int numNearPerfect = 0; for (int n = 1; n <= 10000; n++) { int divisorSum = 0; for (int divisor = 1; divisor < n; divisor++) { if (n % divisor == 0) { divisorSum += divisor; } } if (divisorSum == n) { numPerfect++; println(" " + n + " is perfect."); } else if ((n - divisorSum <= 2) && (divisorSum - n <= 2)) { numNearPerfect++; println(" " + n + " is near perfect."); } } println(); println("There are these many perfect numbers between " +
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14S-Section-Solution - CS106A Handout 14S April 13th 15th,...

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