u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 45773 Accepted Submission(s): 21027
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
题目解析：
这个题的意思就是让你输出n从0到9公式求出的值，是一个非常简单的题，但是要注意输出。
代码：
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int i,j,jiecheng[10];
double a[10];
cout<<"n e"<<endl<<"- -----------"<<endl;
for(i=0; i<=9; i++)
{
for(j=0; j<i; j++)
{
jiecheng[0]=1;
jiecheng[i]=i*jiecheng[i-1];
}
a[0]=1;
a[i]=a[i-1]+1.0/jiecheng[i];
if(i==0) cout<<"0"<<" "<<"1"<<endl;
if(i==1) cout<<"1"<<" "<<"2"<<endl;
if(i==2) cout<<"2"<<" "<<"2.5"<<endl;
if(i>=3)
{
cout<<i<<" ";
printf("%.9lf\n",a[i]);
}
}
return 0;
}
