n个骰子的点数(递归) 代码(C)
本文地址: http://blog.csdn.net/caroline_wendy
题目: 把n个骰子仍在地上, 全部骰子朝上一面的点数之和为s. 输入n, 打印出s的全部可能的值出现的概率.
採用递归的方法, 能够如果仅仅有一个骰子, 然后骰子数递增相加.
代码:
/*
* main.cpp
*
* Created on: 2014.7.12
* Author: spike
*/ #include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h> using namespace std; const int g_maxValue = 6; 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);
}
}
} void Probability (int number, int* pProbabilities) {
for (int i=1; i<=g_maxValue; ++i)
Probability(number, number, i, 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;
} int main(void)
{
PrintProbability(2);
return 0;
}
输出:
2: 2.777778e-002
3: 5.555556e-002
4: 8.333333e-002
5: 1.111111e-001
6: 1.388889e-001
7: 1.666667e-001
8: 1.388889e-001
9: 1.111111e-001
10: 8.333333e-002
11: 5.555556e-002
12: 2.777778e-002