## EMC Interview Question for Software Engineer / Developers

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0
of 0 vote

use random number generation; then %10; then 0~4 -> A, 5~8 -> B, 9 -> C

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0
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int method()
{
for(int i=1;i<=10;i++)
{
if(i<=4)return a;
elseif(i<=9)return b;
else return c;
}
}

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0

hehe, wise guy

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0
of 0 vote

Given K discrete events with different probabilities P[k], produce a random value k consistent with its probability.

The obvious way to do this is to preprocess the probability list by generating a cumulative probability array with K+1 elements:

``````C = 0
C[k+1] = C[k]+P[k].``````

Note that this construction produces C[K]=1. Now choose a uniform deviate u between 0 and 1, and find the value of k such that C[k] <= u < C[k+1]. Although this in principle requires of order \log K steps per random number generation, they are fast steps, and if you use something like \lfloor uK \rfloor as a starting point, you can often do pretty well.

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0
of 0 vote

``````char method()
{
int x=(int)Math.random*9;
if(x<=3)
return 'a';
if(x<=8)
return 'b';
return 'c';
}``````

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0
of 0 vote

``````char method()
{
int x=(int)Math.random()*9;
if(x<=3)
return 'a';
if(x<=8)
return 'b';
return 'c';
}``````

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0
of 0 vote

dude wht a dumass , after return the stack will be deallocated (google it)

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0

you missed to see the if statement before return statement.

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0
of 0 vote

``````import java.util.Random;

/**
*
*/

/**
* @author Lazycoder
*
*/
public class Percentage {

/**
* @param args
*/
public static void main(String[] args) {
Random rand = new Random();
int i = rand.nextInt(10);
if(i < 4){
System.out.println("A");
}else if(i < 9){
System.out.println("B");
}else{
System.out.println("C");
}

}

}``````

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-1
of 1 vote

``````int method()
{
return 40 % A, 50 % B && 10 % C
}``````

????

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0

Haha awesome! :)

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-1
of 1 vote

``````for(int i=1;i<=10;i++)
{
if (i<5) return 'A';
else if i(<9) return 'B';
else return return 'C';
}``````

Name:

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