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C++实现N个骰子的点数算法

2020-01-26 15:17:58
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本文实例讲述了C++实现N个骰子的点数算法,分享给大家供大家参考之用。具体方法如下:

题目要求:把n个骰子仍在地上,所有点数

实现代码如下:

#include <iostream>using namespace std;const int g_maxValue = 6;const int number = 6;int array[(number - 1) * g_maxValue + 1];void probility(int original, int current, int sum, int *array){ if (current == 0) { array[sum - original]++; return; } for (int i = 1; i <= g_maxValue; i++) { probility(original, current - 1, sum + i, array); }}void generateValue(int *array, int size){ if (array == NULL || size <= 0) return; for (int i = number; i <= number * g_maxValue; i++) { array[i - number] = 0; } probility(number, number, 0, array); for (int i = 0; i < (number - 1) * g_maxValue + 1; i++) { cout << array[i] << " "; }}void main(){ generateValue(array, (number - 1) * g_maxValue + 1);}

采用循环解法,和上述代码本质一样,但是略有不同。

具体实现代码如下:

#include <iostream>using namespace std;void PrintProbability(int number){ const int g_maxValue = 6; int *array[2]; for (int i = 0; i < 2; i++) { array[i] = new int[g_maxValue * number]; } for (int i = 0; i < g_maxValue * number; i++) { array[0][i] = 0; array[1][i] = 0; }  int flag = 0; for (int i = 0; i < g_maxValue; i++) { array[flag][i] = 1; } for (int i = 1; i < number; i++) { for (int j = 0; j < i; j++)  array[1 - flag][j] = 0; for (int j = i; j < (i + 1) * g_maxValue; j++) {  array[1 - flag][j] = 0;  for (int k = 1; k <= j && k <= g_maxValue; k++)  array[1 - flag][j] += array[flag][j - k]; } flag = 1 - flag; } for (int i = number - 1; i < g_maxValue * number; i++) { cout << array[flag][i] << " "; }}void main(){ PrintProbability(2);}

希望本文所述对大家C++程序算法设计的学习有所帮助。

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