作業三：代碼複審

   我審查的劉彥麟同學的代碼，我已經用CFREE運作，可以成功運作也發符合條件。并且每一行代碼都很清晰，可以很清晰的看出每一行代碼的意義，有很好的規範性，在複查過程中，我也學習了很多，包括對代碼的規範性。我覺得代碼複查時件很有用，很能學東西的事情！

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int add(int num1,int num2)
{
 return num1+num2;
}
int subtraction(int num1,int num2)
{
 return num1-num2;
}
int mul(int num1,int num2)
{
 return num1*num2;
}
float divis(int num1,int num2)
{
 return ((int)(((float)num1/num2)*100+0.5))/100.0;
}
int main()
{ 
 int type = -1;
 int data = -1;
 int choice,num1,num2,results;
 float div_result;
 char ch;
 int i,j,num3,num4; 
 printf("請輸入四則運算題目的數量：\n");
 scanf("%d",&j); 
 srand((unsigned int)time(NULL));
 for(i=0;i<j;i++){
     num3=rand()%100+1;
     num4=rand()%100+1;
       choice=rand()%8;
  switch (choice)
  {
  case 1:
    num1 = rand()%100+1;
    num2 = rand()%100+1;
    printf("%d+%d=\n",num1,num2);
   break;
  case 2:
    num1 = rand()%100+1;
    num2 = rand()%100+1;
    printf("%d-%d=\n",num1,num2);
   break;
  case 3:
    num1 = rand()%10;
    num2 = rand()%10;
    printf("%d*%d=\n",num1,num2);
   break;
  case 4:
    num1 = rand()%10;
    num2 = rand()%10+1;
    printf("%d/%d=\n",num1,num2);
   break;
    case 5:
    num1 = rand()%100+1;
    num2 = rand()%100+1;
    printf("%d/%d+%d/%d=\n",num1,num3,num2,num4);
   break;
  case 6:
    num1 = rand()%100+1;
    num2 = rand()%100+1;
    printf("%d/%d-%d/%d=\n",num1,num3,num2,num4);
   break;
  case 7:
    num1 = rand()%10;
    num2 = rand()%10;
    printf("%d/%d*%d/%d=\n",num1,num3,num2,num4);
   break;
  case 8:
    num1 = rand()%10;
    num2 = rand()%10+1;
    printf("%d/%d/%d/%d=\n",num1,num3,num2,num4);
   break;
  }
}
}