Singleton having a has relation; predict the output.

Greedy algorithms for Job Scheduling

In this programming problem and the next you'll code up the greedy algorithms from lecture for minimizing the weighted sum of completion times.. This file describes a set of jobs with positive and integral weights and lengths. It has the format [number_of_jobs] [job_1_weight] [job_1_length] [job_2_weight] [job_2_length] ... For example, the third line of the file is "74 59", indicating that the second job has weight 74 and length 59. You should NOT assume that edge weights or lengths are distinct. Your task in this problem is to run the greedy algorithm that schedules jobs in decreasing order of the difference (weight - length). This algorithm is not always optimal. IMPORTANT: if two jobs have equal difference (weight - length), you should schedule the job with higher weight first. Beware: if you break ties in a different way, you are likely to get the wrong answer. You should report the sum of weighted completion times of the resulting schedule --- a posi...

Integer comparison using == gets tricky between -128 to 127.

Integer i=45; Integer k=45; System.out.println("i.hashcode= "+i.hashCode()); System.out.println("k.hashcode= "+k.hashCode()); System.out.println(i==k); System.out.println(i.equals(k)); Out Put: i.hashcode= 45 k.hashcode= 45 true true Great, it seems work fine! now change the value of i=k=201. Integer i=201; ...

You are given four integers 'a', 'b', 'y' and 'x', where 'x' can only be either zero or one. Your task is as follows: If 'x' is zero assign value 'a' to the variable 'y', if 'x' is one assign value 'b' to the variable 'y'. You are not allowed to use any conditional operator (including ternary operator). Possible Solutions: 1. y = (1 - x) * a + x * b; 2.y = a ^ ((a ^ b) & (x))

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