the sum of 1 a a^2 _ _ _ a^n-CodePudding

here is my code:

#include <iostream>
using namespace std;

int main()
{
    int s = 1, i, n, m;
    double a;
    cin >> a >> n;
    while (n <= 0)
    {
        cin >> n;
    }
    m = a;
    for (i = 1; i <= n; i  )
    {
        s  = m;
        m *= a;
    }
    cout << s << endl;
    return 0;
}

here was the start data: Given a real number a and a natural n, calculate the sum 1 a a^2 _ _ _ a^n, without using the formula for the sum of a geometric progression. The running time of the program must be proportional to n.

Input data Enter 2 numbers - a and n.

Output It is necessary to display the value of the sum. The checking program considers my code not full. Please explain where is the problem?

CodePudding user response：

Firstly, you don't need this while loop:

while (n <= 0)
{
    std::cin >> n;
}

What it does is keep input into n until you input a positive number, which probably isn't what you want.

Second, m *= a;. m is an int type, while a is a double type

#include <iostream>
// using namespace std; is bad, so don't use it

int main()
{
    // there's no need to declare a variable but leave it uninitialized.
    int a{};
    int n{};

    std::cin >> a >> n;

    unsigned long long m{a}; // this series grows very fast
                             //so it's best to store it in unsigned long long type
    unsigned long long s{1}  // same thing happen for s
    for (int i = 1; i <= n; i  ) // declare i in the loop scope
    {
        s  = m;
        m *= a;
    }                            // so now i is destroyed
    std::cout << s << '\n'; // use newline character instead of std::endl
    return 0;
}

CodePudding user response：

You are using an infinite loop to read the value of n.

#include <iostream>
using namespace std;

int main()
{
int s = 1, i, n, m;
double a;
cin >> a >> n;
// The below code was causing an endless loop
// while (n <= 0)
// {
//     cin >> n;
// }
m = a;
for (i = 1; i <= n; i  )
{
    s  = m;
    m *= a;
}
cout << s << endl;
return 0;
}

Commenting/removing the part of the code provides the result for the program.