CareerCup

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

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

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] = 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.

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

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

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

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");
		}
		
	}

}

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

????

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