n个骰子的点数

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/***********************************************************************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。***********************************************************************//*解题思路1：有题目可知，最小点数之和为n，最大点数之和为6n。要想求出n个骰子的点数和，可以先把n个骰子分为两堆：第一堆只有一个，另一个有n-1个。单独的那一个可能出现从1到6的点数。我们需要计算从1到6的每一种点数和剩下的n-1个骰子来计算点数和。接下来把剩下的n-1个骰子还是分成两堆，第一堆只有一个，第二堆有n-2个。我们把上一轮那个单独骰子的点数和这一轮单独骰子的点数和相加，再和剩下的n-2个骰子来计算点数和。所有我们可以用递归来解决这个问题。*/#include<stdio.h>#include<math.h>int g_maxValue = 6;void Probability(int number, int* pProbabilities);void printProbability(int number){if(number<1)return;int maxSum = number * g_maxValue;int* pProbabilities = new int[maxSum - number + 1];for(int i=number; i<=maxSum; ++i)pProbabilities[i-number] = 0;Probability(number,pProbabilities);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 original, int current, int sum, int* pProbabilities);void Probability(int number, int* pProbabilities){for(int i=1; i<=g_maxValue; ++i)Probability(number,number,i,pProbabilities);}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);}}}/*解题思路2：由于第一种解法是基于递归实现，有很多计算是重复的。我们可以用两个数组来存储骰子点数的每一个总数出现的次数。在一次循环中，第一个数组中的第n个数字表示骰子和为n的次数。在下一循环中，我们加上一个新的骰子，此时和为n的骰子出现的次数应该等于上一次循环中骰子点数和为n-1,n-2,n-3,n-4.n-5.n-6的次数总和，然后我们把另一个数组的第n个数字设为前一个数组对应的第n-1,n-2,n-3,n-4.n-5.n-6之和。*/// ====================方法二====================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)         pProbabilities[flag][i] = 1;         for (int k = 2; k <= number; ++k)     {        for(int i = 0; i < k; ++i)            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];        }         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];}void test(){printProbability(1);}int main(){test();return 0;}
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