// 面試題60：n個骰子的點數
// 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。

#include <iostream>
#include <math.h>

int g_maxValue = 6;

// ====================方法一====================
//使用遞歸，還是會有重復計算
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];//建立一個長為maxSum - number + 1的數組，用來統計次數
    for (int i = number; i <= maxSum; ++i)//初始化為0
 

        pProbabilities[i - number] = 0;

    Probability(number, pProbabilities);//統計次數

    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);
}

//劃分為1個和n-1個兩堆色子，然後叠代如此劃分，遍歷所有可能，然後pProbabilities數組相應加1
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====================
//使用循環方法，需要找到統計新的一個色子的規律
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)//建立兩個數組，初始化為0
    {
        pProbabilities[0][i] = 0;
        pProbabilities[1][i] = 0;
    }

    int flag = 0;
    for (int i = 1; i <= g_maxValue; ++i)
        pProbabilities[flag][i] = 1;//第一個色子，每個值出現次數為1，值1~g_maxValue

    for (int k = 2; k <= number; ++k)//從第二個色子開始統計
    {
        for (int i = 0; i < k; ++i)
            pProbabilities[1 - flag][i] = 0;//把另一個數組的k之前的次數清零，因為那是以前的統計結果，後面不可能出現的值

        for (int i = k; i <= g_maxValue * k; ++i)//對於另一個數組而言，統計一個新的色子，其每個和的次數等於當前數組的前六個值的和
        {
            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];
}

// ====================測試代碼====================
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;
}

《劍指offer》第六十題（n個骰子的點數）