如何计算获得等于用户输入的随机数并计算 C 中的平均值所需的迭代次数？

Question

使用嵌套循环，我们必须创建一个程序，从用户那里获取 0-99 范围内的数字，并使用种子随机数生成器来尝试猜测数字。在生成用户号码之前，我们必须跟踪 RNG 生成了多少号码。我们还必须执行此过程 50 次，然后计算 RNG 猜测我们的数字所需的平均尝试次数。这是我到目前为止所拥有的一切，我很难过：

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

int main()
{
    srand(time(NULL));

    int userInput;
    int randoms = rand() % 100;
    int numCalls;
    int i;
    float average = 0.0;

    printf("Please enter a number in between 0 and 99: ");
    scanf(" %d", &userInput);

    for( i = 0; i < 50; i++)
    {

        while (userInput != randoms);
        {

           numCalls = numCalls + 1;

           if (userInput = randoms)
           {
               float average = numCalls/50.0;

           }
        }
        printf("Number of iterations it took for a matching number: %d\n", numCalls);
        printf("Average number of iterations to find a match: %.2f\n", average);

    }

return;
}

Answer 1

没有 randoms 值的更新
if (userInput=randoms)不是布尔函数，应该是if (userInput==randoms)而不是

这是我的建议：

int main()
{
    srand(time(NULL));

    int userInput;
    int randoms = rand() % 100;
    int numCalls;
    int i;
    float average = 0.0;

    int totalCalls=0;

    printf("Please enter a number in between 0 and 99: ");
    scanf(" %d", &userInput);

    for( i = 0; i < 50; i++)
    {
        numCalls = 0; //reset iteration number
        do{
            randoms = rand()%100;
            numCalls++
        } while (randoms!=userInput);

        totalCalls += numCalls;         
        printf("Number of iterations it took for a matching number: %d\n", numCalls);
        average = (float)totalCalls/(float)i;
        printf("Average number of iterations to find a match: %.2f\n", average);
    }

return;
}

Answer 2

在这里，您从用户那里获取 1 个输入，并且只计算该输入所需的 RNG 调用次数。将此数字除以 50 不会得到平均值。

您想输入 50 次，即 scanf 语句应该在循环内。如果 numCalls[i] 是第 i 个输入所需的随机数生成器调用次数，则平均值是所有 50 个输入的总和除以 50。

Answer 3

除了更新randoms和相等比较的技术故障外，您必须验证所有用户输入，并且 validate 提供的值在您的预期范围内。否则，如果发生 matching 或 input 失败，您可能会调用 Undefined Behavior。

scanf 有一个 return。您必须使用它来验证 格式字符串 中包含的每个 转换说明符 的成功转换（例如 "%d" 是 1转换）。如果转换成功，则必须验证该值是否介于 0 - 99 之间。您可以通过以下方式完成：

#define MAXN   99
...
    if (scanf ("%d", &num) != 1) {  /* validate ALL user input */
        fprintf (stderr, "error: invalid input.\n");
        return 1;
    }
    if (num < 0 || num > MAXN) {    /* validate num is within range */
        fprintf (stderr, "error: value out of range: %d\n", num);
        return 1;
    }

考虑需要变量的范围。是的，您可以将它们全部声明在顶部，但如果它们仅在循环等中使用，那么如果您在需要的范围内声明值，则可以减少名称冲突或稍后无意中修改值的可能性。（对于 C99 或更高版本，您可以在循环声明中声明 for 循环变量（例如 for (int i = 0; ...)）

将这些部分放在一起，您可以通过验证输入来防止 UB，如上所述，并使用类似以下内容整理变量声明的范围：

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

#define NTIMES 50   /* if you need a constant, declare one (or two) */
#define MAXN   99

int main (void) {

    int i, num;
    double avg = 0.0;

    srand (time (NULL));

    printf ("enter a number (0-%d): ", MAXN);
    if (scanf ("%d", &num) != 1) {  /* validate ALL user input */
        fprintf (stderr, "error: invalid input.\n");
        return 1;
    }
    if (num < 0 || num > MAXN) {    /* validate num is within range */
        fprintf (stderr, "error: value out of range: %d\n", num);
        return 1;
    }

    for (i = 0; i < NTIMES; i++) {  /* loop 50 time for average */
        int nrands = 0,
            randnum = rand() % (MAXN + 1);    /* get 1st random */
        /* loop until equal updating counter nrands */
        for (; randnum != num; randnum = rand() % (MAXN + 1))
            nrands += 1;
        printf ("test[%2d] : %4d iterations to match.\n", i+1, nrands); 
        avg += nrands;      /* keep sum in avg */
    }
    avg /= (double)NTIMES;  /* divide for final average */

    printf ("\nAverage number of iterations to find match: %.2f\n", avg);

    return 0;
}

例子Use/Output

$ ./bin/iters2match
enter a number (0-99): 31
test[ 1] :   19 iterations to match.
test[ 2] :   30 iterations to match.
test[ 3] :  239 iterations to match.
test[ 4] :  116 iterations to match.
test[ 5] :  220 iterations to match.
test[ 6] :  198 iterations to match.
test[ 7] :   64 iterations to match.
test[ 8] :   77 iterations to match.
test[ 9] :  106 iterations to match.
test[10] :   94 iterations to match.
test[11] :  123 iterations to match.
test[12] :   77 iterations to match.
test[13] :  167 iterations to match.
test[14] :   23 iterations to match.
test[15] :    3 iterations to match.
test[16] :  226 iterations to match.
test[17] :   58 iterations to match.
test[18] :  190 iterations to match.
test[19] :   44 iterations to match.
test[20] :  204 iterations to match.
test[21] :  134 iterations to match.
test[22] :   30 iterations to match.
test[23] :   40 iterations to match.
test[24] :   59 iterations to match.
test[25] :   21 iterations to match.
test[26] :  218 iterations to match.
test[27] :    8 iterations to match.
test[28] :   21 iterations to match.
test[29] :  259 iterations to match.
test[30] :  227 iterations to match.
test[31] :   11 iterations to match.
test[32] :   22 iterations to match.
test[33] :  187 iterations to match.
test[34] :   90 iterations to match.
test[35] :    5 iterations to match.
test[36] :   43 iterations to match.
test[37] :  114 iterations to match.
test[38] :   38 iterations to match.
test[39] :   24 iterations to match.
test[40] :   53 iterations to match.
test[41] :   71 iterations to match.
test[42] :  148 iterations to match.
test[43] :   61 iterations to match.
test[44] :   78 iterations to match.
test[45] :    5 iterations to match.
test[46] :   30 iterations to match.
test[47] :  281 iterations to match.
test[48] :   18 iterations to match.
test[49] :  109 iterations to match.
test[50] :   31 iterations to match.

Average number of iterations to find match: 94.28

检查一下，如果您还有其他问题，请告诉我。

如何计算获得等于用户输入的随机数并计算 C 中的平均值所需的迭代次数？

how to count how many iterations it takes to get a random number equal to user input and calculate that average in C?

c

random

nested-loops