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.
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<stdio.h>
using namespace std;
int main()
{
int i,sum;
double a[100];
sum=1;
for(i=0;i<=9;i++)
{
if(i==0)
a[i]=1;
else
{
sum=sum*i;
a[i]=a[i-1]+1.0/sum;
}
}
printf("n e\n- -----------\n");
for(i=0;i<10;i++)
{
cout<<i<<" ";
if(i<2)
printf("%.0lf\n",a[i]);//小数点后保留0位有效数值
else if(i==2)
printf("%.1lf\n",a[i]);//小数点后保留1位有效数值
else
printf("%.09lf\n",a[i]);//小数点后保留9位有效数值
}
return 0;
}
#include<stdio.h>
using namespace std;
int main()
{
int i,sum;
double a[100];
sum=1;
for(i=0;i<=9;i++)
{
if(i==0)
a[i]=1;
else
{
sum=sum*i;
a[i]=a[i-1]+1.0/sum;
}
}
printf("n e\n- -----------\n");
for(i=0;i<10;i++)
{
cout<<i<<" ";
if(i<2)
printf("%.0lf\n",a[i]);//小数点后保留0位有效数值
else if(i==2)
printf("%.1lf\n",a[i]);//小数点后保留1位有效数值
else
printf("%.09lf\n",a[i]);//小数点后保留9位有效数值
}
return 0;
}