Question

在这里compsci新手。我应该编程2个用户输入的值，并在不使用*和/运算符的情况下对它们进行乘法和除法。

我已经想出如何繁殖但不分裂......

    //Multiplication
    cin >> digit1;
    cin >> digit2;

    float product = 0;

    for (int i = 0; i < digit2; i++) product = product + digit1;
    cout << product << endl;

    return 0;

至于师我不太确定......

    cin >> digit1;
    cin >> digit2;

    float quotient = 0;

    //Enter Division operation
    cout << quotient << endl;
    return 0;

感谢您的帮助！

Answer 1

对于除法，可以做类似的事情（下面是digit2 / digit1）：

int movingNum = 0;
int i = 0;
while(movingNum < digit2){
    // This is the counter of how many times digit1 goes into digit2, increment it
    // because the running sum is still less than digit2
    i++;

    // increment the moving sum by digit1 so that it is larger by digit1 moving through
    // the next iteration of the loop
    movingNum += digit1;
}

cout << "digit1 goes into digit2 " << i << " times.";

对于digit1 / digit2：

int movingNum = 0;
int i = 0;
while(movingNum < digit1){
    // This is the counter of how many times digit2 goes into digit1, increment it
    // because the running sum is still less than digit1
    i++;

    // increment the moving sum by digit2 so that it is larger by digit2 moving through
    // the next iteration of the loop
    movingNum += digit2;
}

cout << "digit2 goes into digit1 " << i << " times.";

这些显然是整数除法，如果两个输入不均衡，则会有余数。在上述循环之后，可以通过以下方式计算余数：

int remainder = movingNum - digit2;

如果你真的在寻找一个浮点答案/结果，这将是一个完全不同的答案。

Answer 2

重复划分减去。

示例：4/2字面意思是多少个2构成4（即商）。因此，用除数减去除数直到除数变为零，同时保持从除数（这里为4）减去除数（这里为2）的次数。

尝试以下代码（这不是浮点答案。如果您要这样做，请告诉我！：D）。

#include<stdio.h>

int main() {

    int divi, divs, quo = 0, rem = 0;

    scanf("%d %d", &divi, &divs);

    for(rem = divi - divs; rem >= 0; rem = rem-divs) {

        if(rem < 0)                         // to ensure that the right answer is obtained when the divisor isn't a factor of dividend.
          break;
        else
          quo++;
  }

printf("%d\n", quo);

return 0;

}

如何在不使用除法运算符的情况下除以两个整数？

2 个答案: