[英]Can someone abuse LINQ and solve this money puzzle?
为了它的乐趣,我想看到有人使用和滥用LINQ来解决这个钱的问题。 我真的不知道你会怎么做 - 我想填充一些然后选择它。
如果给出一定数量的硬币并将所有硬币的总金额加在一起:显示每种可能的硬币组合。 硬币是四分之一(.25),尺寸(.10)镍(.05)和便士(.01)
包括选项,以便硬币类型可以为零,或者每个硬币必须至少有1个。
示例问题:如果我有19个硬币并且硬币加起来为1.56美元并且每种类型的硬币必须至少有1个。
答案是:
1个季度,9个尺寸,8个镍,1个便士
2个宿舍,5个角,11个镍,1个便士
2个季度,9个尺寸,2个镍,6个便士
3个宿舍,1个角,14个镍,1个便士
3个宿舍,5个角,5个镍,6个便士
4个宿舍,1个角钢,8个镍币,6个便士
5个季度,1个尺寸,2个镍,11个便士
如果我们允许零投注,我们允许额外获得0个季度,13个角钱,5个镍币,1个便士
下面是一些使用强力方法解决问题的示例C#代码。 不要费心改进样本,让我们看看使用Linq的解决方案。 //如果可能的话,尽量不要使用任何regualar c#循环代码。
private void SolveCoinProblem(int totalNumberOfCoins, double totalAmount, int minimumNumberOfEachCoin)
{
int foundCount = 0;
long numberOfTries = 0;
Console.WriteLine(String.Format("Solving Coin Problem:TotalNumberOfCoins={0}TotalAmount={1}MinimumNumberOfEachCoin{2}", totalNumberOfCoins, totalAmount, minimumNumberOfEachCoin));
for (int totalQuarters = minimumNumberOfEachCoin; totalQuarters < totalNumberOfCoins; totalQuarters++)
{
for (int totalDimes = minimumNumberOfEachCoin; totalDimes < totalNumberOfCoins; totalDimes++)
{
for (int totalNickels = minimumNumberOfEachCoin; totalNickels < totalNumberOfCoins; totalNickels++)
{
for (int totalPennies = minimumNumberOfEachCoin; totalPennies < totalNumberOfCoins; totalPennies++)
{
numberOfTries++;
if (totalQuarters + totalDimes + totalNickels + totalPennies == totalNumberOfCoins)
{
if (Math.Round((totalQuarters * .25) + (totalDimes * .10) + (totalNickels * .05) + (totalPennies * .01),2) == totalAmount)
{
Console.WriteLine(String.Format("{0} Quarters, {1} Dimes, {2} Nickels, {3} Pennies", totalQuarters, totalDimes, totalNickels, totalPennies));
foundCount++;
}
}
}
}
}
}
Console.WriteLine(String.Format("{0} Combinations Found. We tried {1} combinations.", foundCount, numberOfTries));
}
未经测试,但是:
int minQuarters = 1, minDimes = 1,
minNickels = 1, minPennies = 1,
maxQuarters = 19, maxDimes = 19,
maxNickels = 19, maxPennies = 19,
coinCount = 19, total = 156;
var qry = from q in Enumerable.Range(minQuarters, maxQuarters)
from d in Enumerable.Range(minDimes, maxDimes)
from n in Enumerable.Range(minNickels, maxNickels)
from p in Enumerable.Range(minPennies, maxPennies)
where q + d + n + p == coinCount
where q * 25 + d * 10 + n * 5 + p == total
select new {q,d,n,p};
foreach (var row in qry)
{
Console.WriteLine("{0} quarter(s), {1} dime(s), {2} nickel(s) and {3} pennies",
row.q, row.d, row.n, row.p);
}
实际上,出于零售目的 - 或许更好的询问是“我能发出的最少的硬币是什么”? 用。。。来代替:
...
from p in Enumerable.Range(minPennies, maxPennies)
where q + d + n + p <= coinCount
where q * 25 + d * 10 + n * 5 + p == total
orderby q + d + n + p
...
并使用First()
或Take(...)
;-p
您还可以通过在maxDimes
测试中减去(例如) q
来减少已检查案例的数量(等等......) - 类似于(简化):
int minCount = 1,
coinCount = 19, total = 156;
var qry = from q in Enumerable.Range(minCount, coinCount - (3 * minCount))
where q * 25 <= total
from d in Enumerable.Range(minCount, coinCount - (q + (2 * minCount)))
where q * 25 + d * 10 <= total
from n in Enumerable.Range(minCount, coinCount - (q + d + minCount))
where q * 25 + d * 10 + n * 5 <= total
from p in Enumerable.Range(minCount, coinCount - (q + d + n))
where q + d + n + p == coinCount
where q * 25 + d * 10 + n * 5 + p == total
select new { q, d, n, p };
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