c - 查找 3 个骰子之和等于整数 k 的概率时的分割错误错误

Question

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

待办事项：实现下面名为 cum_prob 的函数。 此函数将整数 k 和长整数试验作为输入。 此函数返回双精度值。 在功能中，我们多次掷骰子 3 个骰子。投掷次数是试验。 我们计算 3 个骰子的结果加起来至少为 k 的次数。 然后我们用这个数字和试验来计算概率 3 个骰子的总和至少为 k。 最后，我们返回此概率。

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) = %.5lfn", k, cum_prob(k, n));
    return 0;
}

Answer 1

以下建议的代码：

1. 干净地编译
2. 执行所需的功能
3. 包含一个固定表(在代码中计算(
4. 不使用递归或长循环

现在，建议的代码：

#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( "%dn", i );
        printf("P(sum of the 3 dice is at least %d) = %.5lfnn", i, cum_prob(i, n));
    }
    return 0;
}

以下是有效输入范围的结果

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

您可以将原始源用于 main() 函数，因此它仅在每次运行代码时计算单个用户输入的值。