题目：把n个骰子扔在地上，所有骰子朝上一面的点数之和为S。输入n，打印出S的所有可能的值出现的概率.

递归解法

<pre><code>`
func probility(diceCount:Int) {

    let total:Double = pow(6.0, Double(diceCount))
    let maxValue:Int = 6 * diceCount
    let len:Int = maxValue - diceCount + 1
    var arr:[Int] = [Int].init(repeating: 0, count: len)
    probilityNumber(diceCount: diceCount, arr: &arr)
    for i in 0..<len {
        let result:Double = Double(arr[i]) / total
        print("FlyElephant--\(i+diceCount)---出现的概率----\(result)")
    }
}


func probilityNumber(diceCount:Int,arr:inout [Int]) {
    for i in 1...6 {
        baseProbilityNumber(diceCount: diceCount, remainderDice: diceCount, sum: i, arr: &arr)
    }
}

// diceCount 🎲总数 remainderDice 剩余的🎲
func baseProbilityNumber(diceCount:Int,remainderDice:Int,sum:Int,arr:inout [Int])  {
    if remainderDice == 1 {
        arr[sum - diceCount] = arr[sum - diceCount] + 1
    } else {
        for i in 1...6 {
            baseProbilityNumber(diceCount: diceCount, remainderDice: remainderDice - 1, sum: sum + i, arr: &arr)
        }
    }
}`</code></pre>

非递归解法

<pre><code>`
func diceProbility(diceCount:Int) {
var arr:[[Int]] = []
let maxLen:Int = 6 * diceCount + 1
let first:[Int] = [Int].init(repeating: 0, count: maxLen)
let next:[Int] = [Int].init(repeating: 0, count: maxLen)
arr.append(first)
arr.append(next)

    var flag:Int = 0
    for i in 1...6 {
        arr[flag][i] = 1
    }
    
    
    for i in 2...diceCount {
        for j in 0..<i {
            arr[1-flag][j] = 0
        }
        for k in i...6 * i {
            var m:Int = 1
            arr[1-flag][k] = 0
            while m <= k && m <= 6 {
                arr[1-flag][k] += arr[flag][k-m]
                m += 1
            }
        }
        flag = 1 - flag
    }
    let total:Double = pow(6.0, Double(diceCount))
    for i in diceCount...6 * diceCount {
        let rate:Double = Double(arr[flag][i])/total
        print("FlyElephant--\(i)出现的概率---\(rate)")
    }
}`</code></pre>

测试代码:
<pre><code>var dices:Dices = Dices() dices.probility(diceCount: 3) dices.diceProbility(diceCount: 3)</code></pre>