Question

我正在尝试为我的C类编写一个程序，该程序可以跟踪在银行中的存款。它为您提供一个菜单，其中包含以下选项：输入存款，显示所有存款的总和，显示从最高到最低的存款（使用冒泡排序），显示平均存款，显示最低的存款，然后选择退出选项。据我所知，输入，求和和退出选项工作正常，但其他三个选项已损坏。当您选择它们时，无论您对数组进行了什么输入，其作用都好像它们都等于零。这是我到目前为止所拥有的：

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



int main()

{
    int sortCount, sortCount2, sortCount3, swap;// variables for sort
    int depositCount = 0, sumCount, lowestCount;
    int averageCount, avgSum = 0, avg; //variables for average
    char switchInput = 0;//menu input
    double deposits[100] = { 0 }, sum = 0, average;

    do 
    {
        printf("BANKING MAIN MENU\n\n");
        printf("[M]ake a new deposit\n");
        printf("[S]um of all deposits\n");
        printf("[D]eposits will be displayed from highest to lowest\n");
        printf("[A]verage of all deposits\n");
        printf("[L]owest deposit will be displayed\n");
        printf("[Q]uit\n\n");
        printf("Please enter selection:\n\n");
        scanf(" %c", &switchInput);

        switch (switchInput)
        {
        case 'm': case 'M'://Deposit screen

            printf("\nPlease enter deposit:\n\n");
                scanf("%lf", &deposits[depositCount++]);//deposit input
                ;

            for (sortCount = 0; sortCount < depositCount; sortCount++)//Should sort the array highest to lowest
                for (sortCount2 = 0; sortCount2 < depositCount - sortCount - 1; sortCount2++)
                    if (deposits[sortCount] < deposits[sortCount+1])
                    {
                        swap = deposits[sortCount];
                        deposits[sortCount] = deposits[sortCount+1];
                        deposits[sortCount+1] = swap;
                    }

                break;

        case 's': case 'S'://Total of deposits screen

            for (sumCount = 0; sumCount < depositCount; sumCount++)//depositCount should have it only use parts of the array where there are inputs.
                sum = sum + deposits[sumCount];
                printf("\nYour total deposits equal $%.2lf\n\n", sum);

                break;

        case 'd': case 'D'://Highest to lowest screen


            for (sortCount3 = 0; sortCount3 < depositCount; sortCount3++)//depositCount should have it only use parts of the array where there are inputs.
            {
                printf("$%d \n", deposits[sortCount3]);
            }
            break;

        case 'a': case 'A'://Average screen

            for (sumCount = 0; sumCount < depositCount; sumCount++) //depositCount should have it only use parts of the array where there are inputs.
        {
                avgSum = avgSum + deposits[sumCount];
                avg = avgSum / depositCount;
            }
            printf("Your average deposit is $%.2lf.\n", avg);
            break;

        case 'l': case 'L'://Lowest screen

            printf("The lowest deposit is $%.2lf.\n", deposits[depositCount]);//If the numbers are ordered from highest to lowest, the then entry in the array at the position equal to the number of deposits should be the lowest

            break;

        case 'q': case 'Q'://quit screen

            printf("\nThank you for using our bank!\n\n");
            system("pause");

            return 0;
            break;

        default ://invalid option

            printf("\nInvalid selection!\n\n");
        }

    } while (switchInput != 'q'||'Q');

}

Answer 1

气泡排序未在c中正确排序数组

在

silly
[In]:
silly['priors_inc'] = silly[['thing', 'cond']].where(silly['cond'] == 1).groupby('thing').cumsum() - 1

[Out]:

silly
   thing  cond  priors  priors_inc
0      a     1       0         0.0
1      b     2       0         NaN
2      a     1       1         1.0
3      c     2       0         NaN
4      b     1       0         0.0
5      c     2       0         NaN
6      c     1       0         0.0
7      a     2       2         NaN
8      a     1       2         2.0
9      b     2       1         NaN
10     c     1       1         1.0
11     a     2       3         NaN

for (sortCount = 0; sortCount < depositCount; sortCount++)//Should sort the array highest to lowest for (sortCount2 = 0; sortCount2 < depositCount - sortCount - 1; sortCount2++) if (deposits[sortCount] < deposits[sortCount+1]) { swap = deposits[sortCount]; deposits[sortCount] = deposits[sortCount+1]; deposits[sortCount+1] = swap; }在内部 for 中是未使用的，您可以始终独立地执行相同的操作。此外，您在最后一个索引之后走了1个索引

在S.O上有很多气泡排序的实现。我让你搜索并更正

交换必须为 double

在

sortCount2

除法必须在求和之后进行，每次都不要，所以

case 'a': case 'A'://Average screen
       for (sumCount = 0; sumCount < depositCount; sumCount++) //depositCount should have it only use parts of the array where there are inputs.
   {
           avgSum = avgSum + deposits[sumCount];
           avg = avgSum / depositCount;
       }

和 avgSum 和 avg 必须为 double

   case 'a': case 'A'://Average screen
        for (sumCount = 0; sumCount < depositCount; sumCount++) //depositCount should have it only use parts of the array where there are inputs.
        {
            avgSum = avgSum + deposits[sumCount];
        }
         avg = avgSum / depositCount;

必须

while (switchInput != 'q'||'Q');

气泡排序未在C中正确排序数组

1 个答案: