~~~ # u Calculate e Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 37341????Accepted Submission(s): 16897 ~~~ 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 題目大意為：求出0-9的情況下這個公式求出來的e的值是多少，從結果可以看出來，0和1的時候保留整數部分，2的時候保留一位小數，3,4以及以后的時候保留九位小數。 ~~~ ~~~ 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); } } } ~~~