n个骰子的点数

   计算机作为一种工具，它的作用是用来解决实际的生产生活中的问题。程序员的工作就是把各种现实问题抽象成数学模型并用计算机的编程语言表达出来。建立模型:第一步:选择合理的数据结构表达问题；第二步:分析模型中的内在规律，并且用编程语言表达这种规律。

问题: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。

    详细的实现代码如下:

int g_maxValue = 6;  //易于程序的扩展，比如设计点数最大为12的骰子
 
// ====================方法一:基于递归求骰子点数，时间效率不高====================
void Probability(int number, int* pProbabilities);
void Probability(int original, int current, int sum, int* pProbabilities);

void PrintProbability_Solution1(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;
 
    Probability(number, pProbabilities);
 
    int total = pow((double)g_maxValue, number);
    for(int i = number; i <= maxSum; ++i)
    {
        double ratio = (double)pProbabilities[i - number] / total;
        printf("%d: %e\n", i, ratio);
    }
 
    delete[] pProbabilities;
}
 
void Probability(int number, int* pProbabilities)
{
    for(int i = 1; i <= g_maxValue; ++i)
        Probability(number, number, i, pProbabilities);
}
 
void Probability(int original, int current, int sum, 
                 int* pProbabilities)
{
    if(current == 1)
    {
        pProbabilities[sum - original]++;
    }
    else
    {
        for(int i = 1; i <= g_maxValue; ++i)
        {
            Probability(original, current - 1, i + sum, pProbabilities);
        }
    }
} 

// ====================方法二:基于循环求骰子点数，时间效能好====================
void PrintProbability_Solution2(int number)
{
    if(number < 1)
        return;

    int* pProbabilities[2];
    pProbabilities[0] = new int[g_maxValue * number + 1];
    pProbabilities[1] = new int[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 = 0; i < k; ++i)
            pProbabilities[1 - flag][i] = 0;

        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 = (double)pProbabilities[flag][i] / total;
        printf("%d: %e\n", i, ratio);
    }
 
    delete[] pProbabilities[0];
    delete[] pProbabilities[1];
}

// ====================测试代码====================
void Test(int n)
{
    printf("Test for %d begins:\n", n);
    
    printf("Test for solution1\n");
    PrintProbability_Solution1(n);

    printf("Test for solution2\n");
    PrintProbability_Solution2(n);

    printf("\n");
}

int _tmain(int argc, _TCHAR* argv[])
{
    Test(1);
    Test(2);
    Test(3);
    Test(4);
    
    Test(11);

    Test(0);

    return 0;
}