JZ-C-43

Posted 2020-07-29 回看欧洲

剑指offer第四十三题：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

  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 }