c#中的增量谜题

本文关键字： | 更新日期: 2023-09-27 18:04:21

在比赛中，您使用以下策略下注。每次你赌输了，下一轮的赌注就翻倍。无论何时你赢了，下一轮的赌注将是1美元。你赌1美元开始。

例如，如果你一开始有20元，你在第一轮中赢了，在接下来的两轮中输了，然后在第四轮中赢了，那么你最终会得到20+1-1-2+4 = 22元。

您需要完成函数getFinalAmount，它接受两个参数:

第一个参数是一个整数initialAmount，它是我们开始投注时的初始金额。
第二个参数是字符串betResults。结果的第i个字符将是"W"(赢)或"L"(输)，表示第i轮的结果。你的函数应该返回所有回合结束后你将拥有的钱数。

如果在某一时刻你的账户中没有足够的钱来支付下注的价值，你必须停止并归还你在那个时刻的钱。

我试过这段代码，失败了:

var amountInHand = 15;
var possiblities = "LLLWLLLL";
static int Calculate(int amountInHand, string possibles)
{
    var lastBet = 1;
    foreach (char c in possiblities)
    {
        if (c == 'W')
        {
            amountInHand = amountInHand + 1;
        }
        else if (c == 'L')
        {
            var loss = 0;
            if (lastBet == 1)
            {
                loss = lastBet;
            }
            else if (lastBet > 1)
            {
                loss = lastBet * 2;
            }
            amountInHand = amountInHand - loss;
            lastBet++;
        }
    }
    return amountInHand;
}

预期输出

1st round - Loss: 15-1 = 14
2nd round - Loss: 14-2 = 12 (Bet doubles)
3rd round - Loss: 12-4 = 8
4th round - Win: 8 + 8 = 16
5th round - Loss:16-1 = 15 (Since the previous bet was a win, this bet has a value of 1 dollar)
6th round - Loss: 15-2 = 13
7th round - Loss: 13-4 = 9
8th round - Loss: 9-8 = 1

c#中的增量谜题

这是R.B.给出的正确答案，但不知道他为什么删掉了。

      var amountInHand = 15;
        var possiblities = "LLLWLLLL";
        var lastBet = 1;

        foreach (char c in possiblities)
        {
            if (c == 'W')
            {
                amountInHand = amountInHand + lastBet;
                lastBet = 1;
            }
            else if (c == 'L')
            {
                amountInHand = amountInHand - lastBet;
                lastBet = lastBet * 2;
            }
            //handle running out of money
            if (lastBet > amountInHand) return amountInHand;
        }

关于这个问题，我写了一篇博文和一段代码高尔夫。我的答案是:

return possibilities.Aggregate(new{b=1,c=amountInHand,x=1},(l,o)=>l.x<0?l:o=='W'?new{b=1,c=l.c+l.b,x=1}:new{b=l.b*2,c=l.c-l.b,x=(l.c-l.b)-(l.b*2)}).c;

Un-golfed:

private static int getFinalAmount(string q, int w)
{
    return possibilities.Aggregate(new { b = 1, c = amountInHand, x = false }, //initial seed gets the flag for cancel set to false
                    (l, o) =>
        l.x     //if our cancel flag is set, 
            ? l      //just return the same result
            : o == 'W'   //if the outcome was a win
                        //do the math and now also set a cancel flag to false (if we won, we can make our next bet for sure)
                ? new { b = 1, c = l.c + l.b, x = false }
                        //do our math again, but this time the cancel flag is tricky.  
                : new { b = l.b * 2, c = l.c - l.b,
                                //we cancel if our new amount will be less than our new bet.  
                                //Note, we can't use the new values that we just set in the same section - 
                                //they're not available yet so we have duplicate math here.
                            x = (l.c - l.b) < (l.b * 2) })
                    .c;   //all the function returns is the current amount
}