# JZ-C-43

  1 //============================================================================
2 // Name        : JZ-C-43.cpp
3 // Author      : Laughing_Lz
4 // Version     :
5 // Copyright   : All Right Reserved
6 // Description : n个骰子的点数：把n个骰子扔在地上，所有骰子朝上一面的点数之和为s，求所有可能的s的概率
7 //============================================================================
8
9 #include <iostream>
10 #include <math.h>
11 #include <stdio.h>
12 using namespace std;
13
14 int g_maxValue = 6; //骰子面数
15
16 // ====================方法一====================
17 void Probability(int number, int* pProbabilities);
18 void Probability(int original, int current, int sum, int* pProbabilities);
19 /**
20  * 递归：许多计算时重复的，当number变大时，性能很低
21  */
22 void PrintProbability_Solution1(int number) {
23     if (number < 1)
24         return;
25
26     int maxSum = number * g_maxValue;
27     int* pProbabilities = new int[maxSum - number + 1];
28     for (int i = number; i <= maxSum; ++i)
29         pProbabilities[i - number] = 0;
30
31     Probability(number, pProbabilities);
32
33     int total = pow((double) g_maxValue, number);//pow(x,y)函数：表示x^y
34     for (int i = number; i <= maxSum; ++i) {
35         double ratio = (double) pProbabilities[i - number] / total;
36         printf("%d: %e\n", i, ratio);
37     }
38
39     delete[] pProbabilities;
40 }
41
42 void Probability(int number, int* pProbabilities) {
43     for (int i = 1; i <= g_maxValue; ++i)
44         Probability(number, number, i, pProbabilities);
45 }
46
47 void Probability(int original, int current, int sum, int* pProbabilities) {
48     if (current == 1) {
49         pProbabilities[sum - original]++;
50     } else {
51         for (int i = 1; i <= g_maxValue; ++i) {
52             Probability(original, current - 1, i + sum, pProbabilities);
53         }
54     }
55 }
56
57 // ====================方法二====================
58 /**
59  * 循环：利用两个数组交替存储两次循环后各值可能出现的次数。时间性能好
60  */
61 void PrintProbability_Solution2(int number) {
62     if (number < 1)
63         return;
64
65     int* pProbabilities[2];
66     pProbabilities[0] = new int[g_maxValue * number + 1];
67     pProbabilities[1] = new int[g_maxValue * number + 1];
68     for (int i = 0; i < g_maxValue * number + 1; ++i) {
69         pProbabilities[0][i] = 0;
70         pProbabilities[1][i] = 0;
71     }
72
73     int flag = 0;
74     for (int i = 1; i <= g_maxValue; ++i)
75         pProbabilities[flag][i] = 1;//第一次,先将各值可能出现的次数赋为1
76
77     for (int k = 2; k <= number; ++k) {//number 骰子数
78         for (int i = 0; i < k; ++i)
79             pProbabilities[1 - flag][i] = 0;
80
81         for (int i = k; i <= g_maxValue * k; ++i) {
82             pProbabilities[1 - flag][i] = 0;//先重新赋值为0 ★★(结合flag交替使用得知：这里每一轮都会重复计算上一轮得过的次数，因为每次都是从值为0开始。所以仍有重复计算的问题)
83             for (int j = 1; j <= i && j <= g_maxValue; ++j)
84                 pProbabilities[1 - flag][i] += pProbabilities[flag][i - j];//★★★这里表示f(n)=f(n-1)+f(n-2)+f(n-3)+f(n-4)+f(n-5)+f(n-6)
85         }
86
87         flag = 1 - flag;//交替使用两个数组
88     }
89
90     double total = pow((double) g_maxValue, number);
91     for (int i = number; i <= g_maxValue * number; ++i) {
92         double ratio = (double) pProbabilities[flag][i] / total;
93         printf("%d: %e\n", i, ratio);
94     }
95
96     delete[] pProbabilities[0];
97     delete[] pProbabilities[1];
98 }
99
100 // ====================测试代码====================
101 void Test(int n) {
102     printf("Test for %d begins:\n", n);
103
104     printf("Test for solution1\n");
105     PrintProbability_Solution1(n);
106
107     printf("Test for solution2\n");
108     PrintProbability_Solution2(n);
109
110     printf("\n");
111 }
112
113 int main(int argc, char** argv) {
114     Test(1);
115     Test(2);
116     Test(3);
117     Test(4);
118     Test(11);
119     Test(0);
120
121     return 0;
122 }

—————————————————————————————————————行走在人猿的并行线——Laughing_Lz
posted @ 2016-06-24 20:04  回看欧洲  阅读(161)  评论(0编辑  收藏  举报