《剑指offer》第六十题（n个骰子的点数）

// 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 = 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;
}