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
题目大意为:求出0-9的情况下这个公式求出来的e的值是多少,从结果可以看出来,0和1的时候保留整数部分,2的时候保留一位小数,3,4以及以后的时候保留九位小数。
import java.text.DecimalFormat;
public class Main{
public static void main(String[] args) {
double sum = 0;
System.out.println("n e");
System.out.println("- -----------");
for(int i = 0; i < 10; i++)
{
sum += 1.0 / functionMuti(i);
if( i == 0 || i == 1)
{
DecimalFormat decimalForm = new DecimalFormat("0");
System.out.println(i + " " + decimalForm.format(sum));
}
else if(i == 2)
{
DecimalFormat decimalForm = new DecimalFormat("0.0");
System.out.println(i + " " + decimalForm.format(sum));
}
else
{
DecimalFormat decimalForm = new DecimalFormat("0.000000000");
System.out.println(i + " " + decimalForm.format(sum));
}
}
}
private static int functionMuti(int i) {
if(i == 1)
return 1;
else if(i == 0)
return 1;
else
{
return i * functionMuti(--i);
}
}
}
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