发现的3个骰子的总和的一个等于整数k的概率时分段错误

Question

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

TO DO：实现cum_prob下面命名的功能。此函数采用的整数k，以及一个长整型试验作为输入。这个函数返回一个双精度值。在温控功能，我们折腾了3个骰子多次。投掷的数量是试验。我们计算的次数是3个骰子的结果加起来至少为K。然后我们使用这个号码和试验，以计算出的3个骰子的总和至少k的概率。最后，我们回到这个概率看起来。

double cum_prob(int k, long trials)
{   
    double count = 0;
    double all_trials = 0;
    double prob;
    if (trials == 0)
        return prob;
    if (rand() % 18 + 3 == k)
{
        (count ++);
        (all_trials ++);
        return cum_prob(k, -- trials);
}
    else
        {
            (all_trials ++);
            return cum_prob(k, -- trials);
        }
    prob = ((count / all_trials) * 100);


}


//Do not change the following code.
int main()

    long n = 10000000;
    int k;

    printf("Enter k :");    
    scanf("%d", &k);
    assert(k>= 3 && k<=18);
    srand(12345);
    printf("P(sum of the 3 dice is at least %d) = %.5lf\n", k, cum_prob(k, n));
    return 0;

}

Answer 1

下面提出的代码：

1. 完全编译
2. 执行期望的功能
3. 采用了固定台（在代码计算）
4. 不使用递归，也不会长久循环

现在，所提出的代码：

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


double cum_prob(int targetValue, long trials)
{
    (void)trials;

    int possibleResults[19] = {0};

    // build table
    for( int dice1 = 1; dice1 <=6; dice1++ )
    {
        for( int dice2 = 2; dice2 <= 6; dice2++ )
        {
            for( int dice3 = 1; dice3 <= 6; dice3++ )
            {
                possibleResults[ dice1+dice2+dice3 ]++;
            }
        }    
    }

    // calculate total possible 
    double totalPossibilites = 0.0;
    for( int index=0; index<=18; index++ )
    {    
        totalPossibilites += (double)possibleResults[index];
    }

    int sumLessEqualTarget = 0;
    for( int i = 0; i<targetValue; i++ )
    {
        sumLessEqualTarget += possibleResults[i];
    }


    return  100.0 - ((double)sumLessEqualTarget / totalPossibilites) *100.0;
}


//Do not change the following code   
int main()
{
    long n = 10000000;
    //int k;

    //printf("Enter k :");    
    //scanf("%d", &k);
    //assert(k>= 3 && k<=18);
    //srand(12345);
    for( int i = 3; i<=18; i++ )
    {
        printf( "%d\n", i );
        printf("P(sum of the 3 dice is at least %d) = %.5lf\n\n", i, cum_prob(i, n));
    }
    return 0;
}

这里是有效的输入范围的结果

3
P(sum of the 3 dice is at least 3) = 100.00000

4
P(sum of the 3 dice is at least 4) = 100.00000

5
P(sum of the 3 dice is at least 5) = 99.44444

6
P(sum of the 3 dice is at least 6) = 97.77778

7
P(sum of the 3 dice is at least 7) = 94.44444

8
P(sum of the 3 dice is at least 8) = 88.88889

9
P(sum of the 3 dice is at least 9) = 80.55556

10
P(sum of the 3 dice is at least 10) = 69.44444

11
Enter k P(sum of the 3 dice is at least 11) = 56.66667

12
P(sum of the 3 dice is at least 12) = 43.33333

13
P(sum of the 3 dice is at least 13) = 30.55556

14
P(sum of the 3 dice is at least 14) = 19.44444

15
P(sum of the 3 dice is at least 15) = 11.11111

16
P(sum of the 3 dice is at least 16) = 5.55556

17
P(sum of the 3 dice is at least 17) = 2.22222

18
P(sum of the 3 dice is at least 18) = 0.55556

您可以使用原来的源为main()功能，因此它仅在代码的每次运行计算单个用户输入的值。