HDU 1012 u Calculate e （水题）

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

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
double ans,sum;
printf("n e\n");
printf("- -----------\n");
printf("0 1\n");
ans=sum=1;
for(int i=1;i<=9;i++) {
ans*=i;
sum+=1.0/ans;
if(i==1) printf("%d %0.0lf",i,sum);
else if(i==2) printf("%d %0.1lf",i,sum);
else printf("%d %0.9lf",i,sum);
printf("\n");
}
return 0;
}