C++问题 - 有人可以帮忙吗？

发布于 2024-10-20 13:42:19 字数 1005 浏览 7 评论 0原文

你介意帮我解决以下问题吗？谢谢。

问题： 一次性掷出第 6 面 3 个骰子的百分比可以通过数学计算或模拟。蒙特卡罗方法是一个计算机过程，用于找出问题的解决方案通过计算机模拟问题。编写一个程序，掷三个骰子，计算它们的值求和，并找出滚动每个可能结果的概率。

给定一个骨架程序 q1dsculpture.c，它生成滚动 a 的统计信息六面骰子 10000 次。修改程序，使其生成掷出三个六面骰子的总和的统计数据。程序输出的示例在下列的。请注意，由于骰子滚动的随机性。

提示：每次掷骰子时调用 rand() 三次，每个骰子调用一次。

骨架：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define SIZE 7

int main() {

  int face, roll, frequency[SIZE] = { 0 };

  srand( time(NULL) );

  for (roll = 1; roll <= 10000; roll++) {
    face = rand() % 6 + 1;
    ++frequency[face];
  }
  printf("%s%12s\n", "Face", "Frequency");

  for (face = 1; face <= SIZE - 1; face++)
    printf("%4d%12d\n", face, frequency[face]);
  getchar();
}

输出：

Face Frequency
3     49
4     129
5     276
6     481
7     669
8     994
9     1131
10    1213
11    1269
12    1197
13    962
14    707
15    464
16    268
17    144
18    47

原文

Would you mind helping me to solving the following question? Thanks.

Question:
The percentage of rolling 3 dices of face 6 in one go can be found out mathematically or
simulation. The Monte Carlo method is a computer process to find out the solution of a
problem by computer simulation. Write a program that roll three dices, calculate their
sum, and find out the probability of rolling each possible outcome.

You are given a skeleton program q1dskeleton.c that generates the statistics of rolling a
dice of six sides 10000 times. Modify the program so that it generates the statistics of the sum of rolling three six-sided dices. An example of the program output is given in the
following. Note that because of the random nature of dice rolling.

Hint: Call rand() three times for every throw of the dices, one for each dice.

Skeleton:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define SIZE 7

int main() {

  int face, roll, frequency[SIZE] = { 0 };

  srand( time(NULL) );

  for (roll = 1; roll <= 10000; roll++) {
    face = rand() % 6 + 1;
    ++frequency[face];
  }
  printf("%s%12s\n", "Face", "Frequency");

  for (face = 1; face <= SIZE - 1; face++)
    printf("%4d%12d\n", face, frequency[face]);
  getchar();
}

output:

Face Frequency
3     49
4     129
5     276
6     481
7     669
8     994
9     1131
10    1213
11    1269
12    1197
13    962
14    707
15    464
16    268
17    144
18    47

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抹茶夏天i‖ 2024-10-27 13:42:19

显然，您必须增加 SIZE 因为不再有 6 种可能性，而是 18 种 - 我将在此处包括 1 和 2 的不可能总数，因为您的原始解决方案包括不可能的值 0 :-)

然后，您不再扔一个骰子，而是扔三个，然后将骰子的值相加。所以类似于（伪代码）：

dice1 = first random number
dice2 = second random number
dice3 = third random number
face = dice1 plus dice2 plus dice3

加法就是您用来增加特定数组元素的内容。

为了获得奖励积分，您应该避免打印出不可能的结果。这是对最终 for 语句起始位的相当简单的修改。

顺便说一下，您期望的比率是 {1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1}因此“理想”输出应该相当接近：（

Face   Frequency
   3          46
   4         138
   5         277
   6         462
   7         694
   8         972
   9        1157
  10        1250
  11        1250
  12        1157
  13         972
  14         694
  15         462
  16         277
  17         138
  18          46

尽管这只加起来只有 9992 个样本 - 至于其他 8 个样本出现在哪里，这是一个冒险）。

Obviously, you have to increase SIZE since there are no longer 6 possibilities, but 18 - I'll include the impossible totals of 1 and 2 here since your original solution included the impossible value of zero :-)

Then, instead of throwing one dice, you throw three and then add up the values. So something like (pseudo-code):

dice1 = first random number
dice2 = second random number
dice3 = third random number
face = dice1 plus dice2 plus dice3

That addition is what you then use to increase a specific array element.

For bonus points, you should probably avoid printing out the impossible outcomes. That's a fairly simple modification to the starting bit of your final for statement.

The ratios you expect, by the way, are {1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1} so the "ideal" output should be reasonably close to:

Face   Frequency
   3          46
   4         138
   5         277
   6         462
   7         694
   8         972
   9        1157
  10        1250
  11        1250
  12        1157
  13         972
  14         694
  15         462
  16         277
  17         138
  18          46

(even though that only adds up to 9992 samples - it's a crapshoot as to where the other eight show up).

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