// 面试题60：n个骰子的点数// 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。#include 
     
      #include 
      
       int g_maxValue = 6;// ====================方法一====================//使用递归，还是会有重复计算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];//建立一个长为maxSum - number + 1的数组，用来统计次数    for (int i = number; i <= maxSum; ++i)//初始化为0        pProbabilities[i - number] = 0;    Probability(number, pProbabilities);//统计次数    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);}//划分为1个和n-1个两堆色子，然后迭代如此划分，遍历所有可能，然后pProbabilities数组相应加1void 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====================//使用循环方法，需要找到统计新的一个色子的规律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)//建立两个数组，初始化为0    {        pProbabilities[0][i] = 0;        pProbabilities[1][i] = 0;    }    int flag = 0;    for (int i = 1; i <= g_maxValue; ++i)        pProbabilities[flag][i] = 1;//第一个色子，每个值出现次数为1，值1~g_maxValue    for (int k = 2; k <= number; ++k)//从第二个色子开始统计    {        for (int i = 0; i < k; ++i)            pProbabilities[1 - flag][i] = 0;//把另一个数组的k之前的次数清零，因为那是以前的统计结果，后面不可能出现的值        for (int i = k; i <= g_maxValue * k; ++i)//对于另一个数组而言，统计一个新的色子，其每个和的次数等于当前数组的前六个值的和        {            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 main(int argc, char* argv[]){    Test(1);    Test(2);    Test(3);    Test(4);    Test(11);    Test(0);    system("pause");    return 0;}