(1). 求:1*2*3*4*5.......*n的值
- package suanfa;
- /**
- * Created by tl on 2016/4/10.
- */
- public class Digui {
- public static int digui(int n){
- if(n==1||n==0){
- return n;
- }else{
- System.out.println("执行第" + n + "次");
- return n*digui(n-1);
- }
- }
- public static void main (String[] args){
- System.out.print(digui(n));
- }
(2). 求1+2+3+4+5......+n的值
- static int count(int n){
- if(n>0){
- return n+count(n-1);
- }else{
- return 0;
- }
- }
- public static void main(String args[])
- {
- int sum=count(n);
- System.out.println(sum);
- }
- }
(3).求1,1,2,3,5,8,13,21,34.....n位的值
- static int count(int n){
- if(n==1||n==2) {
- return 1;
- }
- return count(n-1)+count(n-2);
- }
- public static void main(String args[])
- {
- int sum=count(n);
- System.out.println(sum);
- }