《剑指offer》第六十题（n个骰子的点数）

《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0pProbabilities[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: %en", 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_maxValuefor (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: %en", 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 solution1n");PrintProbability_Solution1(n);printf("Test for solution2n");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;
}

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