Question

我的编程课程的练习要求我们为收银员编写一个程序，提示应付金额和收到的金额并计算更改。它还必须计算所需的美元，四分之一，硬币，硬币和便士的数量。它工作正常，直到达到硬币，镍币和硬币。这就是输出变得不稳定的地方。我还注意到visual studio读取0.10为.1000000000000000056。镍和硬币也是如此，但美元或季度不会发生这种情况。我已经尝试了几种不同的方法来解决它，但没有运气。任何建议

cout << "Amount Due. \n";
double due;
cout << "$";
    cin >> due;

cout << "Amount recieved. \n";
double rec;
cout << "$";
    cin >> rec;

double change = (rec - due);
cout << "Total change= $" << change << " \n";

int dollars = change / 1.00;
change = change - dollars;

int quarters = change / .25;
change = change - quarters;

int dimes = change / .10;
change = change - dimes;

int nickels = change / .05;
change = change - nickels;

int pennies = change / .01;
change = change - pennies;

cout << "Number of dollars: " << dollars << " \n";
cout << "Number of quarters: " << quarters << " \n";
cout << "Number of dimes: " << dimes << " \n";
cout << "Number of nickels: " << nickels << " \n";
cout << "Number of pennies: " << pennies << " \n";

return 0;

}

Answer 1

while(change > 0){
    if(change >= 1.00){
        dollars ++;
        change -= 1;
        continue;
    }
    if(change >= 0.25){
        quarters ++;
        change -= 0.25;
        continue;
    }
    if(change >= 0.10){
        dimes ++;
        change -= 0.1;
        continue;
    }
    if(change >= 0.05){
        nickels ++;
        change -= 0.05;
        continue;
    }
    if(change >= 0.01){
        pennies ++;
        change -= 0.01;
        continue;
    }
}

continue关键字从顶部再次启动循环。此实现可抵抗浮点不准确性。

返回更改程序小数c ++

1 个答案: