阿姆斯壮数C++

如何询问用户是否要继续

如果正确或错误：

输出如下所示：

请输入一个数字以检查是否为阿姆斯特朗号码：371 371 是阿姆斯特朗号码，是否继续？再见

如果写“Y”，它将在同一个问题中继续

我工作：

while (y != 0)
{
    d=y%10;
    z=z+(d*d*d);
    y=y/10;
}
if ( z == x )
{
 cout <<x <<" is an Armstrong Number" << endl;
}
if ( z != x )
{
cout <<x << " is not an Armstrong Number"<<endl;
}

cout << "Do you want to continue (Y/N) ? " ;
    cin >> m;
    do{
    cout <<"Please enter a number to check if it is an Armstrong number:";
cin >> x;
    if ( z == x )
{
 cout <<x <<" is an Armstrong Number" << endl;
}
if ( z != x )
{
cout <<x << " is not an Armstrong Number"<<endl;
}
cout << "Do you want to continue (Y/N) ? " ;
    cin >> m;
    }while ( m == 'Y' || m == 'y') ;

 cout << "Bye"<<endl;

---

但这是错误的

在声明变量后，启动do while循环：do{；其中“ans”是存储用户答案的char变量。如下所示：

        //Here are the declared variables.
   do{
      cout << "Please enter a number to check if it is an Armstrong number: ";
      cin >> number; //Replace 'number' with the variable name you have used to store that number.
      //The algorithm, which tests if the number is Armstrong number or not, goes here.
      cout << "Do you want to continue (Y/N)? ";
      cin >> ans;
      }while(ans == 'Y');
      cout << "Bye";

---

您可以将整个内容封装在另一个while循环中，切换后，如果输入为n，则中断；我从来没有见过一个开关/箱子被塞进一个应该比这个更糟糕的if/else中你的意思是：如果z==x cout@Lablabla，我做了，但它运行不正确！来自《惠利！》0'直到'cout'

while (y != 0)
{
    d=y%10;
    z=z+(d*d*d);
    y=y/10;
}
if ( z == x )
{
 cout <<x <<" is an Armstrong Number" << endl;
}
if ( z != x )
{
cout <<x << " is not an Armstrong Number"<<endl;
}

cout << "Do you want to continue (Y/N) ? " ;
    cin >> m;
    do{
    cout <<"Please enter a number to check if it is an Armstrong number:";
cin >> x;
    if ( z == x )
{
 cout <<x <<" is an Armstrong Number" << endl;
}
if ( z != x )
{
cout <<x << " is not an Armstrong Number"<<endl;
}
cout << "Do you want to continue (Y/N) ? " ;
    cin >> m;
    }while ( m == 'Y' || m == 'y') ;

 cout << "Bye"<<endl;

        //Here are the declared variables.
   do{
      cout << "Please enter a number to check if it is an Armstrong number: ";
      cin >> number; //Replace 'number' with the variable name you have used to store that number.
      //The algorithm, which tests if the number is Armstrong number or not, goes here.
      cout << "Do you want to continue (Y/N)? ";
      cin >> ans;
      }while(ans == 'Y');
      cout << "Bye";