C 编程作业 - 数学导师 - 除法和余数逻辑 - 我可以用更好的方法来做吗？

Question

所以我为 C 编程作业编写了这个程序。它使用切换菜单条目在加法、减法、乘法和除法中随机生成的问题之间进行选择。

我制作它时遇到的唯一问题是除法部分。如何处理随机生成的不能均匀划分的数字。所以我想，“好吧，甚至不要将小数点作为 float 处理，只需直接整数。然后使用两个随机生成的数字的 mod 来输入余数(如果有的话)。”

这就是我的想法。

void division(){

int a,b;
int ans, uans;
int remain, uremain;

a=rand()%(100-1+1)+1;
b=rand()%(100-1+1)+1;

printf("What is\n");
printf("   %.0f\n",a);
printf(" ÷ %.0f\n",b);
printf("========\n");
printf("?= ");

scanf("%d",&uans);

printf("\n Remainder = ");

scanf("%d",&uremain);

ans = a / b;
remain = a % b;


if(ans==uans && remain==uremain){
    printf("\n\nCorrect!.......\n\n");
    printf("\n\nThe answer was %d with a remainder of %.0f\n",ans, remain);
}
else{
    printf("\n\nIncorrect!......\n\n");
    printf("\n\nThe answer was %d with a remainder of %d\n",ans, remain);
}
}

是否有更好的方法来做到这一点，以解决如果没有余数则可能不必输入余数的问题，我假设我可以执行类似 if(remain > 0) 的操作来查看是否有余数然后添加它，但只是想知道其他人的想法。

对于上下文，这里是完整的程序

/*****************************

CS 50 - Programing in C
Math tutor

Write a program that displays a menu as shown in the sample run. 
You can enter 1, 2, 3, or 4 for choosing an addition, subtraction, multiplication, 
or division test. After a test is finished, the menu is redisplayed. 
You may choose another test or enter 5 to exit the system. 

Each test generates two random single-digit numbers to form a question for addition,
subtraction, multiplication, or division. 

For a subtraction such as number1 – number2,

number1 is greater than or equal to number2.  

For a division question such as number1 / number2, number2 is not zero.

******************************/


#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <ctype.h>
#include <time.h>

int displaymenu();
void addition();
void subtraction();
void multiplication();
void division();


int main(){

int com;
srand(time(NULL));

do{
    com = displaymenu();
    switch(com){
        case 1:
            addition();
            break;
        case 2:
            subtraction();
            break;
        case 3:
            multiplication();
            break;
        case 4:
            division();
            break;
        case 5: printf("Have a nice day\n");
    }
}while(com != 5);   

//needed for all basic programs to run for the professor
system("pause");

return(0);
}

int displaymenu(){

int choice;

printf("*-----------------------*\n");
printf("|       MATH TUTOR      |\n");
printf("*-----------------------*\n");
printf("*    1. Addition        *\n");
printf("*    2. Subtration      *\n");
printf("*    3. Multiplication  *\n");
printf("*    4. Division        *\n");
printf("*    5. EXIT            *\n");
printf("*-----------------------*\n");
printf("Menu Choice: ");
scanf("%d",&choice);
return choice;      
}

void addition(){

int a,b, ans;

a=rand()%(100-1+1)+1;
b=rand()%(100-1+1)+1;

printf("What is\n");
printf("   %d\n",a);
printf(" + %d\n",b);
printf("========\n");
printf("?= ");
scanf("%d",&ans);

if((a+b)==ans){
    printf("\n\nCorrect!.......\n\n");
}
else{
    printf("\n\nIncorrect!......\n\n");
    printf("\n\nThe answer was %d\n",a+b);
}   
}

void subtraction(){

int a,b, ans;

a=rand()%(100-1+1)+1;
b=rand()%(100-1+1)+1;

printf("What is\n");
printf("   %d\n",a);
printf(" - %d\n",b);
printf("========\n");
printf("?= ");
scanf("%d",&ans);

if((a-b)==ans){
    printf("\n\nCorrect!.......\n\n");
}
else{
    printf("\n\nIncorrect!......\n\n");
    printf("\n\nThe answer was %d\n",a-b);
}   
}

void multiplication(){

int a,b, ans;

a=rand()%(100-1+1)+1;
b=rand()%(100-1+1)+1;

printf("What is\n");
printf("   %d\n",a);
printf(" x %d\n",b);
printf("========\n");
printf("?= ");
scanf("%d",&ans);

if((a*b)==ans){
    printf("\n\nCorrect!.......\n\n");
}
else{
    printf("\n\nIncorrect!......\n\n");
    printf("\n\nThe answer was %d\n",a*b);
}   
}

void division(){

int a,b;
int ans, uans;
int remain, uremain;

a=rand()%(100-1+1)+1;
b=rand()%(100-1+1)+1;

ans = a / b;
remain = a % b;

printf("What is\n");
printf("   %d\n",a);
printf(" ÷ %d\n",b);
printf("========\n");
printf("?= ");
scanf("%d",&uans);


if(remain > 0){
    printf("\nRemainder = ");
    scanf("%d",&uremain);
}



if(ans==uans && remain==uremain){
    printf("\n\nCorrect!.......\n");
    printf("\n\nThe answer was %d with a remainder of %d\n",ans, remain);
}
else{
    printf("\n\nIncorrect!......\n");
    printf("\n\nThe answer was %d with a remainder of %d\n\n\n\n",ans, remain);
}   
}

Answer 1

您可以像乘法问题一样生成除法问题，但使用乘积和其中一个数字来解决问题，另一个数字作为解决方案。

也就是说，像乘法一样生成a和b，然后问a*b是什么> 除以 a(这样 b 就是答案)。