43.n个骰子的点数

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解法一：基于递归求骰子点数

void PrintProbability(int number)
{
if (number < 1)
return;
int maxSum = number*g_maxValue;
int* PrintProbability = 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);
}
}
}

解法二：基于循环求骰子点数

void PrintProbability(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 i = k; i <= g_maxValue*k; ++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];
}

43.n个骰子的点数

标签：

原文地址：http://blog.csdn.net/wangfengfan1/article/details/46841777