Question

我创建了一个程序来显示总和并显示一个人输入的反转数字。 sum函数有效，但反转功能不起作用。任何人都可以给我任何关于如何解决它的提示。我创建了一个程序来显示总和并显示一个人输入的反转数字。 sum函数有效，但反转功能不起作用。任何人都可以给我任何关于如何解决它的提示。

  #include<iostream>
    #include<iomanip> 

    using namespace std;

    void printSum(int n, bool reverse);
    int sm(int n);
    int reverseInt(int n);
    void printAddTable(int n);
    int main()
    {   
    int reverse;
       int sum=0;
           int n;
        cout<<"Enter N value:"<<endl;
        cin>>n;

        if(n>0)
        {
           reverse = true;
           printSum( n, reverse);    // calls the printSum Method
        }
        else
        {
          //cout<<"enter positive Number only:"<<endl;
        }
            sum = sm(n);        //err    // calls the sum Method
         reverse = reverseInt(n);    // calls the reverseInt Method
        cout<<"The reverse value is:"<<reverse;
        printAddTable(n);            // calls the printAddTable Method
        //getch()
       }
       //end of main()
     void printSum(int n, bool reverse)
    {

    int sum=0;            
                        // print sum of reverse numbers
    for(int i=n; i>=1; i--)
    {
    sum=sum+i;
    cout<<i<< " "<<"+"<<" ";

    }
    cout<<sum;
    }

    int sm(int n)
    {int sum =0;
    for(int i=1; i<=n; i++)
   {
         sum = sum + i ;
    cout << endl;

    cout<<i<<" "<<"+"<<" "<<endl;            // print n positive integers
    cout << endl;
    }
    cout<< "Are " <<n<< " positive integers"<<endl;
    cout<< "Sum is "<<sum <<endl;





    return sum;
    }

    int reverseInt(int n)
    {
    int reminder=0;

     int sum =0;    

    while(n<=0)
    {

    reminder = n/10;
    sum = (sum * 10) + reminder;        // it returns the reverse number
    n = n % 10;
    }
    return sum;

    }
    void printAddTable(int n)
    {
    for(int i=1; i<=n; i++)
    {
    cout<<i<<endl;
    for(int j=1; j<=n; j++)            // print n X n add table
    {
    cout<<i+j<<endl;    

    }
    }
    }
    {

Answer 1

在您的代码中，将在评论中提到的reverseInt函数中更改以下内容。

while(n<0)
{

//reminder = n/10; should be
 reminder = n%10;
sum = (sum * 10) + reminder;        // it returns the reverse number
//n = n % 10;
//Should be
n = n/10; //or n /= 10;
}

用c ++反转一个数字

1 个答案: