編程演算法 - n個骰子的點數(遞迴) 代碼(C)

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n個骰子的點數(遞迴) 代碼(C)

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

題目: 把n個骰子仍在地上, 所有骰子朝上一面的點數之和為s. 輸入n, 列印出s的所有可能的值出現的機率.

採用遞迴的方法, 可以假設只有一個骰子, 然後骰子數遞增相加.

代碼:

/* * 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;}

輸出:

2: 2.777778e-0023: 5.555556e-0024: 8.333333e-0025: 1.111111e-0016: 1.388889e-0017: 1.666667e-0018: 1.388889e-0019: 1.111111e-00110: 8.333333e-00211: 5.555556e-00212: 2.777778e-002