編程演算法 - n個骰子的點數(非遞迴) 代碼(C)

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n個骰子的點數(非遞迴) 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.

每次骰子的迴圈過程中, 本次等於上一次n-1, n-2, n-3, n-4, n-5, n-6的次數的總和.

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;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;}

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