系列收敛到C++中的数字e

Question

如何计算系列1 + 1/1!+ 1/2!+ 1/3!+ ... + 1/N!在C++中？我有一个大纲:

#include <iostream>
using namespace std;
int main()
{
    int n, i, j, fat;
    float soma = 0.0;
    cin >> n;
    for (i = 1; i <= n; i++)
    {

        fat = 1;
        soma += 1 / fat;
        for (j = 1; j <= n; j++)
        {
            fat *= j;
        }
    }
    cout << soma << endl;
    return 0;
}

Answer 1

保持运行期限并将其添加到结果中:

double result = 1.0;
double term = 1.0;

for (unsigned int i = 1; i != N; ++i)
{
    term /= i;
    result += term;
}

return result;

您可以exp(x)通过一些小修改来计算任何内容.