題目：

把n個骰子扔在地上，所有骰子朝上一面的點數之和為S。輸入n，打印出S的所有可能的值出現的概率。

解法一：遞迴

       玩過麻將的都知道，骰子一共6個面，每個面上都有一個點數，對應的數字是1到 6之間的一個數字。所以，n個骰子的點數和的最小值為n，最大值為6n。因此，一個直觀的思路就是定義一個長度為6n-n的陣列，和為S的點數出現的次數儲存到陣列第S-n個元素裡。另外，我們還知道n個骰子的所有點數的排列數6^n。一旦我們統計出每一點數出現的次數之後，因此只要把每一點數出現的次數除以n^6，就得到了對應的概率。

       該思路的關鍵就是統計每一點數出現的次數。要求出n個骰子的點數和，我們可以先把n

int g_maxValue = 6;

 

void PrintSumProbabilityOfDices_1(int number)

{

    if(number < 1)

        return;

 

    int maxSum = number * g_maxValue;

    int* pProbabilities = new int[maxSum - number + 1];

    for(int i = number; i <= maxSum; ++i)

        pProbabilities[i - number] = 0;

 

    SumProbabilityOfDices(number, pProbabilities);

 

    int total = pow((float)g_maxValue, number);

    for(int i = number; i <= maxSum; ++i)

    {

        float ratio = (float)pProbabilities[i - number] / total;

        printf("%d: %f\n", i, ratio);

    }

 

    delete[] pProbabilities;

}

 

void SumProbabilityOfDices(int number, int* pProbabilities)

{

    for(int i = 1; i <= g_maxValue; ++i)

        SumProbabilityOfDices(number, number, i, 0, pProbabilities);

}

 

void SumProbabilityOfDices(int original, int current, int value, int tempSum, int* pProbabilities)

{

    if(current == 1)

    {

        int sum = value + tempSum;

        pProbabilities[sum - original]++;

    }

    else

    {

        for(int i = 1; i <= g_maxValue; ++i)

        {

            int sum = value + tempSum;

            SumProbabilityOfDices(original, current - 1, i, sum, pProbabilities);

        }

    }

} 
 

     上述演算法當number比較小的時候表現很優異。但由於該演算法基於遞迴，它有很多計算是重複的，從而導致當number變大時效能讓人不能接受。

void PrintSumProbabilityOfDices_2(int number)

{
    if(number < 1) return;
    double* pProbabilities[2];

    pProbabilities[0] = new double[g_maxValue * number + 1];

    pProbabilities[1] = new double[g_maxValue * number + 1];

    for(int i = 0; i < g_maxValue * number + 1; ++i)

    {

        pProbabilities[0][i] = 0;

        pProbabilities[1][i] = 0;

    }

 

    int flag = 0;

    for (int i = 1; i <= g_maxValue; ++i)

        pProbabilities[flag][i] = 1;

      

    for (int k = 2; k <= number; ++k)

    {

        for (int i = k; i <= g_maxValue * k; ++i)

        {

            pProbabilities[1 - flag][i] = 0;
            for(int j = 1; j <= i && j <= g_maxValue; ++j)

                pProbabilities[1 - flag][i] += pProbabilities[flag][i - j];

        }

 

        flag = 1 - flag;

    }

 

    double total = pow((double)g_maxValue, number);

    for(int i = number; i <= g_maxValue * number; ++i)

    {

        double ratio = pProbabilities[flag][i] / total;

        printf("%d: %f\n", i, ratio);

    }

 

    delete[] pProbabilities[0];

    delete[] pProbabilities[1];

}