Question

我写了一个程序来分隔给定数字的数字。当数字由非零组成时它成功分离但是当内部有一个数字0时，它无法识别并且不打印。我该怎么办？我疯了！

#include <stdio.h>
#include <conio.h>

int quotient (int a, int b);
int remaindar (int a, int b);

int main(void) {

int a,b,number,temp=1,divisor=10000;

printf("Enter three integers: ");
scanf("%d %d %d",&a,&b,&number);

printf("a/b is %d , remainder is %d.\n",quotient(a,b),remaindar(a,b));

temp=number;

while (temp>=1){

        if(temp>=divisor){

            printf("%d  ", quotient(temp,divisor)); 
            temp=remaindar(temp,divisor);
            divisor=divisor/10;
        }

        else divisor=divisor/10;

}


getch();    

return 0;   
}

int quotient (int a, int b){

return a/b; 

}

int remaindar (int a, int b){

return a%b;

}

Answer 1

根据您的信息：第3个数字与商和余数无关。您可以简单地从左到右分隔数字，如下所示：（PS。我假设给定6900你希望看到6,9,0,0）

 #include <iostream>
 void getDigits(int number)
{
    int div = 1;
    //find max divisor, i.e., given 6900, divisor 1000
    //this gives information about how may digits the number has
    while (number / div >= 10) {
      div *= 10;
    } 

    //extract digits from left to right
    while (div != 0) //^^pay attention to this condition, not number !=0
    {
        int currDigit = number /div;
        number %= div;  
           //^^you can change the above two lines to 
          //your quotient and remainder function calls
        div /=10;
        std::cout << currDigit << " "; 
    }
}

int main(){
    int number = 6900;
    std::cout << "test case 1 " <<std::endl;
    getDigits(number);
    int number1 = 5067;
    std::cout << "\ntest case 2 " <<std::endl;
    getDigits(number1);
    int number2 = 12345;
    std::cout << "\ntest case 3 " <<std::endl;
    std::getDigits(number2);
    return 0;
}

请勿使用已弃用的getch()。使用上面的代码，您可以看到以下输出：

test case 1
6 9 0 0
test case 2
5 0 6 7
test case 3
1 2 3 4 5

Answer 2

这种情况正在发生，因为您没有考虑temp小于数字和除数的情况，即数字是0.例如，如果初始数字是302，则除数为{ {1}}和temp为10，打印出2：

如何打印内部为零的数字？

2 个答案: