n个骰子的点数的问题解答

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。

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];
 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 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);
  }
 }
} 

// ====================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];
}