oral_quiz->#N个骰子的点数和#-优快云博客

本文介绍了两种概率计算与概率打印算法的实现过程，包括解决方案一和解决方案二，详细解释了算法逻辑、变量定义和核心步骤，适用于理解概率计算与概率打印的相关概念。

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2019独角兽企业重金招聘Python工程师标准>>>

#include <stdio.h>
#include <math.h>

int g_maxValue = 6;

// ====================SolutionI====================
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(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, sum+i, pProbabilities);
}

//=========================SolutionII============================
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;
	}

	//当前辅助数组
	int aidFlag = 0;
	for(int i=1; i<=g_maxValue; ++i)
		pProbabilities[aidFlag][i] = 1;

	for(int curr=2; curr<=number; ++curr) {			//curr:当前骰子颗数
		for(int i=curr; i<=g_maxValue*curr; ++i) {
			pProbabilities[1-aidFlag][i] = 0;
			for(int j=1; j<=g_maxValue && (i-j>=1); ++j)
				pProbabilities[1-aidFlag][i] += pProbabilities[aidFlag][i-j];
		}

		aidFlag = 1 - aidFlag;
	}

	double total = pow((double)g_maxValue, number);
	for(int i=number; i<=g_maxValue*number; ++i) {
		double ratio = (double)pProbabilities[aidFlag][i] / total;
		printf("%d: %e\n", i, ratio);
	}

	delete[] pProbabilities[0];
	delete[] pProbabilities[1];
}



// ====================TEST====================
void Test(int n)
{
    printf("Test for %d begins:\n", n);

//    printf("Test for solution1\n");
//    PrintProbability_Solution1(n);

    printf("Test for solution2\n");
    PrintProbability_Solution2(n);

    printf("\n");
}

int main(int argc, char* argv[])
{
    Test(1);
    Test(2);
    Test(3);
    Test(4);

    Test(11);

    Test(0);

    return 0;
}

转载于:https://my.oschina.net/ITHaozi/blog/282818