编程算法 - n个骰子的点数(非递归) 代码(C)

n个骰子的点数(非递归) 代码(C)

本文地址: http://blog.csdn.net/caroline_wendy

题目: 把n个骰子仍在地上, 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.

每次骰子的循环过程中, 本次等于上一次n-1, n-2, n-3, n-4, n-5, n-6的次数的总和.

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;

void PrintProbability(int number) {
	const int g_maxValue = 6;
	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];
}

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

编程算法 - n个骰子的点数(非递归) 代码(C)