HDU1012

u Calculate e

Problem Description

A simple mathematical formula for e is



where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

 

Output

Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.

 

Sample Output

n e- -----------0 11 22 2.53 2.6666666674 2.708333333

//HDU1012 基础题//u Calculate e//2017.05.24 by wyj#include#includeusing namespace std;int main(){int factorial(int n);int i, n;int f[10];double e[10];for (i = 0;i <= 9;i++)//0到9的阶乘f[i] = factorial(i);for (n = 0;n <= 9;n++)//n取0到9时e的值{e[n] = 0;for (i = 0;i <= n;i++)e[n] += 1.0 / f[i];}cout << "n e" << endl << "- -----------" << endl;cout << "0 1" << endl;cout << "1 2" << endl;cout << "2 2.5" << endl;for (n = 3;n <= 9;n++)cout << n << " " << setiosflags(ios::fixed) << setprecision(9) << e[n] << endl;return 0;}int factorial(int n)//求阶乘{int r = 1;int i;for (i = 1;i <= n;i++)r *= i;return r;}