《剑指offer》第六十题：n个骰子的点数

Posted 2021-03-01 源周率

tags:

篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了《剑指offer》第六十题：n个骰子的点数相关的知识，希望对你有一定的参考价值。

// 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 <cstdio>
#include <math.h>

int g_maxValue = 6;

// ====================方法一====================
void Probability(int current, int sum, int* pProbabilities);

void PrintProbability_Solution1(int number)
{
    if (number < 1)
        return;

    int maxSum = number * g_maxValue; //n个骰子的最大点数
    //注意此时 + 1是因为数组从0计数
    int* pProbabilities = new int[maxSum + 1];  //存放点数和s的出现次数
    for (int i = number; i <= maxSum; ++i)  //为了直观,数组的[0~number-1]没有用
        pProbabilities[i] = 0;

    for (int i = 1; i <= g_maxValue; ++i)  //第一个骰子, i指该骰子的点数
        Probability(number, i, pProbabilities);

    int total = pow((double)g_maxValue, number); //点数所有可能总数
    for (int i = number; i <= maxSum; ++i)
    {
        double ratio = (double)pProbabilities[i] / total;
        printf("%d: %e
", i, ratio);
    }

    delete[] pProbabilities;
}

void Probability(int current, int sum, int* pProbabilities) //更新current个骰子点数和
{
    if (current == 1) 
        pProbabilities[sum]++;
    else
    {
        //f(n,s) = f(n-1,s-1) + f(n-1,s-2) + f(n-1,s-3)
        //       + f(n-1,s-4) + f(n-1,s-5) + f(n-1,s-6)
        for (int i = 1; i <= g_maxValue; ++i)
            Probability(current - 1, i + sum, pProbabilities);  //通过i+sum传递当前点数和
    }
}

// ====================方法二====================
// 原理看懂了, 实现没咋看懂
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)
    {
        pProbabilities[0][i] = 0;
        pProbabilities[1][i] = 0;
    }

    int flag = 0;
    for (int i = 1; i <= g_maxValue; ++i)  //第一个骰子, 每个值只出现一次, 第n个数字表示和为n出现的次数
        pProbabilities[flag][i] = 1;

    for (int k = 2; k <= number; ++k)  //从第二个骰子开始
    {
        for (int i = 0; i < k; ++i)  //清空另一个数组的前k项, 以前的统计结果
            pProbabilities[1 - flag][i] = 0;

        for (int i = k; i <= g_maxValue * k; ++i)
        {
            pProbabilities[1 - flag][i] = 0;
            for (int j = 1; j <= i && j <= g_maxValue; ++j)
                pProbabilities[1 - flag][i] += pProbabilities[flag][i - j];  //前一个数组 n-1 ~ n-6之和
        }

        flag = 1 - flag;  //交换数组
    }

    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
", i, ratio);
    }

    delete[] pProbabilities[0];
    delete[] pProbabilities[1];
}

// ====================测试代码====================
void Test(int n)
{
    printf("Test for %d begins:
", n);

    printf("Test for solution1
");
    PrintProbability_Solution1(n);

    printf("Test for solution2
");
    PrintProbability_Solution2(n);

    printf("
");
}

int main(int argc, char* argv[])
{
    Test(1);
    Test(2);
    Test(3);
    Test(4);

    //Test(11);

    Test(0);

    return 0;
}