The monkey divide the peach Two by the way of recursion

<span style= "FONT-SIZE:18PX;" >package Lianx;


public class Monkeytao {


	static int ts=0;//peach Total
    int fs=1;//record number of times
    static int hs=5;//Monkey number ...
    The range of the number of peaches in int tsscope=5000;//. Too big to overflow.
    public int FT (int t) {
        if (t==tsscope) {
            //) cancels recursive
            System.out.println ("End") when the peach number is the largest range of values;
           return 0;
        } else{
            if ((t-1)%hs==0 && fs <=hs) {
                if (fs==hs) {
                    System.out.println ("Peach number =" +ts + "satisfies the peach condition"); c14/>}
                fs+=1;
                Return ft ((t-1)/5*4); Back to the monkey to take away a copy of the remaining total
            }else{
                //
               The number of times that did not meet the condition fs=1;//score was reset to 1 returns
               ft (ts+=1);// Peaches number plus +1}} public
    static void Main (string[] args) {
        new Monkeytao (). FT (0);
    }
} </span>