Question

这是我想要实现的输出：

（随机化var1）/（随机化var2）=
答：（var ans）

我已经完成了加法，减法和乘法，但是我遇到了除法的困难，因为我需要精确的被除数和除数来划分，这样就不难回答了。

示例：

五分之四十=
答案：8

不是这一个：

7/5 =
ans：浮动值

这是我的代码：

int x,num,num2,ans,quo,score=0;
time_t t;

clrscr();

for(x=0;x<5;x++)
{
    srand((unsigned) time(&t));
    num2=rand()%10;
    quo=num/num2;

    if(num2==1)
    {
        num=rand()%9;
    }
    else if(num2==2)
    {
        num=2 + (2 * rand ()) %18; //this should return a value divisible by 2(ranges 0-18)
    }
    else if(num2==3)
    {
        num=rand()% //this should return a value divisible by 3 only (ranges 0-27)
    }
    else if(num2==4)
    {
        num=rand()% //this should return a value divisible by 4 only (ranges 0-36)
    }
    else if(num2==5)
    {
        num=rand()% //this should return a value divisible by 5 only (ranges 0-45)
    }
    else if(num2==6)
    {
        num=rand()% //this should return a value divisible by 6 only (ranges 0-54)
    }
    else if(num2==7)
    {
        num=rand()% //this should return a value divisible by 7 only (ranges 0-63)
    }
    else if(num2==8)
    {
        num=rand()% //this should return a value divisible by 8 only (ranges 0-72)
    }
    else if(num2==9)
    {
        num=rand()% //this should return a value divisible by 9 only (ranges 0-81)
    }
    else if(num2==10)
    {
        num=rand()% //this should return a value divisible by 10 only (ranges 0-90)
    }
    else
    {
    }

    gotoxy(30,14);
    printf("\n%d / %d = ",num,num2);
    printf("\nAns: ");
    scanf("%d",&ans);
}

Answer 1

您可以随机选择一个结果并创建问题

denominator = 14 (randomly chosen)
result = 21 (randomly chosen)

numerator = denominator * result

然后你问numerator / denominator

多少钱

Answer 2

简单地：

num2=rand()%9+1;
quo=rand()%10;
num = quo * num2;
printf("\n%d / %d = ",num,num2);

此外，您应该在循环之前将srand()调用移至。否则，如果有人过快地回答问题，他们会再次得到同样的问题。

Answer 3

你可以做到

float quo=(float)num/num2;
printf("%f\n", quo);

并打印确切的结果

如果您想要在它们之间可分割的随机数字，那么您需要其他东西。

你也不会在num之前发起num/num2;，而你也不会检查num2是否为零，这样你就可能被抛出。

finaaly你可以做点像

num2=rand()%10;
num=rand();
while((float)num/num2 != (float)(num/num2))
    num=rand();
int quo=num/num2;

如果绝对只能接受可以分居的夫妻

Answer 4

使用模数运算符进行测试：

int test, result;

test = 6%3;  //test == 0
if(test == 0) result = 6/3; //test passes, assignment made

test = 7%3; //test == 1
if(test == 0) result = 7/3; //test fails, assignment not made

这将保证比率产生整数值。

，随机生成器函数可以使事情变得更容易，如下所示：

int randGenerator(int min, int max)
{
    int random, trying;

    trying = 1;         
    while(trying)
    {
        srand(clock());
        random = (rand()/32767.0)*(max+1);
        (random >= min) ? (trying = 0) : (trying = 1);
    }
    return random;
}

Answer 5

答案：

int x，num，num2，ans，quo，score = 0; time_t t;

clrscr（）; srand（（无符号）时间（＆amp; t））;

for(x=0;x<5;x++)
{
num2=rand()%9;

if(num2==1)
{
 srand((unsigned) time(&t));
 num=rand()%9;
}
else if(num2==2)
{
 srand((unsigned) time(&t));
 num=(rand()%9+1)*2;
}
else if(num2==3)
 {
  srand((unsigned) time(&t));
  num=(rand ()%9+1)*3;
 }
else if(num2==4)
 {
  srand((unsigned) time(&t));
  num=(rand()%9+1)*4;
 }
else if(num2==5)
 {
  srand((unsigned) time(&t));
  num=(rand()%9+1)*5;
 }
else if(num2==6)
 {
  srand((unsigned) time(&t));
  num=(rand()%9+1)*6;
 }
else if(num2==7)
 {
  srand((unsigned) time(&t));
  num=(rand()%9+1)*7;
 }
else if(num2==8)
 {
  srand((unsigned) time(&t));
  num=(rand()%9+1)*8;
 }
else if(num2==9)
 {
  srand((unsigned) time(&t));
  num=(rand()%9+1)*9;
 }
else if(num2==10)
 {
  srand((unsigned) time(&t));
  num=(rand()%9+1)*10;
 }
else
 {
  clrscr();
  printf("");
 }

quo=num/num2;
clrscr();
gotoxy(30,14);
printf("\n%d / %d = ",num,num2);
printf("\nAns: ");
scanf("%d",&ans);

}

使用rand（）在除法表中生成数字

5 个答案: