计算彩票赔率 - 关于代码的问题

Question

I was analyzing the following code concerning Lottery Odds and I have 2 questions:

• why do we need to set lotteryOdds' value to 1 first (also, is it a usual situation in calculations of this type?)

• I know the formula used for counting the odds (n*(n-1)...(n-k+1)/(1*2...*k), but if someone explained me how it works in accordance to repeating the "for" loop, I'd appreciate.

   import java.util.Scanner; public class LotteryOdds { public static void main(String[] args) { Scanner in = new Scanner(System.in); System.out.println(" How many numbers do you need to draw? "); int k = in.nextInt(); System.out.println(" what is the highest numbers you can do draw? "); int n = in.nextInt(); int lotteryOdds = 1; for (int i = 1; i <= k; i++) { lotteryOdds = lotteryOdds * (n - i + 1) / i; System.out.println("Your odds are 1 in " + lotteryOdds + " . Good luck!"); } }

  }

Answer 1

The lotteryOdds variable needs to be 1, because if you do 0 * something you will always get 0, so the program will always return 0. As for the loop, lets change the values and go trough the loop: First iteration:

lotteryOdds = 1 * (n - 1 + 1) / 1 = n / 1

Second iteration:

lotteryOdds = n * ((n - 2 + 1) / 2) = (n / 1) * ((n - 1) / 2) = n * ((n - 1) / 2) = ((n * (n - 1)) / (1 * 2)

Etc. So when you generalize it you get: (n * (n - 1)...(n - k + 1) / (1 * 2 ... * k) just like in the formula above.