int g_maxNumber = 6;                                //筛子最大点数为6void Problility(int number){
    if(number<1)
        return;    int maxAllNumbers = 6 * number;    int length = maxAllNumbers - number;      //数组长度    int *problility = new int[length];              //用来记录筛子和为多少的个数    //初始化每个s的概率都为0    for (int i = number; i < length; i++)           problility[i-number] = 0;    findProbility(number, problility);             //核心函数，用来计算每个sum的个数，number <=sum<=maxAllNumbers    //所有组合    int total = pow(g_maxNumber, 6);    //下面计算s为number 到maxAllNumbers 之间分别概率多少    for (int i = number; i <= maxAllNumbers; i++)    {        double ratio = (double)problility[i - number] / total;        printf("和为%d的概率为：    %s", i, ratio);    }}void findProbility(int number,int *probilities)  //number ：筛子的个数,也代表点数最小的时候,probilities保存每个sum的个数{    for (int i = 1; i <= g_maxNumber; i++)        findProbility(number, number, i, probilities);}/* original 就是筛子的个数，current是当前筛子个数，每次递归current就会-1,sum将   每个筛子的点数累积加起来，pProbilities就是一个数组，记录筛子和为sum的概率  */void findProbility(int original, int current, int sum, int *pProbilities){    //出口    if (current == 1)        pProbilities[sum - original]++;  //之所以-original是因为数组的第0个元素=number开始的,original 就是筛子的个数number    //递归：分成两堆，一堆一个筛子，另一堆剩余的筛子    for (int i = 1; i <=g_maxNumber;i++)        findProbility(original, current - 1, sum + i, pProbilities);}