如何在不使用除法運算符的情況下分割兩個整數？

Question

compsci新手在這裏。我應該編寫2個用戶輸入的值，並且不用*和/運算符將它們相乘並分開。

我已經找到了如何繁殖，但不能分割......

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

    float product = 0; 

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

    return 0; 

爲師我不完全相信......

cin >> digit1; 
    cin >> digit2; 

    float quotient = 0; 

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

感謝您的幫助！

Answer 1

對於鴻溝，我們可以做這樣的事情（下面將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."; 

對於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."; 

這些顯然是對整數除法，如果兩個輸入不等分，則會有餘數。這剩餘部分可以在上面的循環後進行計算：

int remainder = movingNum - digit2; 

如果你是真正的尋找一個浮點答案/結果，這將是一個完全不同的答案。