// 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 <iostream>

#include <math.h>

int g_maxValue = ;

// ====================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 < )

        return;

    int maxSum = number * g_maxValue;

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

    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 = ; 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1

void Probability(int original, int current, int sum, int* pProbabilities)

{

    if (current == )

    {

        pProbabilities[sum - original]++;

    }

    else

    {

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

        {

            Probability(original, current - , 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 < )

        return;

    int* pProbabilities[];

    pProbabilities[] = new int[g_maxValue * number + ];

    pProbabilities[] = new int[g_maxValue * number + ];

    for (int i = ; i < g_maxValue * number + ; ++i)//建立两个数组，初始化为0

    {

        pProbabilities[][i] = ;

        pProbabilities[][i] = ;

    }

    int flag = ;

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

        pProbabilities[flag][i] = ;//第一个色子，每个值出现次数为1，值1~g_maxValue

    for (int k = ; k <= number; ++k)//从第二个色子开始统计

    {

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

            pProbabilities[ - flag][i] = ;//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[ - flag][i] = ;

            for (int j = ; j <= i && j <= g_maxValue; ++j)

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

        }

        flag =  - 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[];

    delete[] pProbabilities[];

}

// ====================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();

    Test();

    Test();

    Test();

    Test();

    Test();

    system("pause");

    return ;

}