HDOJ--1012

u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 38457 Accepted Submission(s): 17411



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 1
1 2
2 2.5
3 2.666666667
4 2.708333333

打表+简单的阶乘运算，不多说什么了，直接上代码：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
double fac(int x)
{
int i;
double f=1;

for(i=1;i<=x;i++)
f*=i;

return f;
}
int main()
{
int n;
int i;
printf("n e\n");
printf("- -----------\n");

double e=2.5;
printf("0 1\n1 2\n2 2.5\n");
for(int i=3;i<=9;i++)
{
e+=1/fac(i);
printf("%d %.9lf\n",i,e);
}
return 0;
}