杭电oj 1012

先看题目

u Calculate e

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

Total Submission(s): 38233 Accepted Submission(s): 17314



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

没什么好说的

#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
double answer[10];
double sum=1;
int i;
answer[0]=1;
for(i=1;i<10;i++)
{
sum=sum*i;
answer[i]=answer[i-1]+1/sum;
}
cout<<"n e"<<endl;
cout<<"- -----------"<<endl;
for(i=0;i<10;i++)
{
if(i>2)
cout<<i<<" "<<setprecision(9)<<std::fixed<<answer[i]<<endl;
else
cout<<i<<" "<<answer[i]<<endl;
}
return 0;
}