Java solution to the Monkey Peach Algorithm solution and expand the monkey algorithm to customize the solution

/* Original Problem Monkey: There is a pile of peaches on the beach, five monkeys to score.
The first monkey put the pile of peaches into five, one more, the monkey threw a lot of one into the sea, took a copy.
The second monkey divided the remaining peaches into five, and one more, it also threw more than one into the sea, took a copy,
Third, four, fifth monkeys do this,
Ask:
At least a few peaches on the beach.

Expand the problem monkey: there is a pile of peaches on the beach, there are n monkeys to divide.
The first monkey divided the pile of peaches into N, more k, and this monkey threw a lot of one into the sea, took a copy.
The second monkey to the rest of the peach and flat into N, and more K, it also put more k into the sea, took a copy,
That's what the last monkey did.
Ask:
At least a few peaches on the beach. The first traditional problem is that the minimum should be 6, with 6 as the beginning of the base in reverse order to find the minimum number of peaches when not meet the conditions of the peach number plus 5 to expand the problem of the smallest solution left should be n+k, In order to find the minimum number of peaches when the condition is not satisfied, make the peach count plus n directly.

*/

public static Class get{
The traditional monkey is divided into peach
public static int Get (int times, int sum, int lastsum) {
0 is returned
if (times = = 0) {
return sum;
}else if (times = = 5) {
Start the calculation again to determine if you want to reset the data
if ((sum-1)%5==0&&sum>=6) {
Return to get (Times-1, Sum, (sum-1)/5*4);
}else {
Return to get (5, 6, 0);
}
}else {
Calculate whether to satisfy this one time peach
if ((lastSum-1)%5==0&&lastsum>6) {
Return to get (Times-1, Sum, (lastSum-1)/5*4);
}else return get (5,sum+5,0);
}
}
Custom monkeys include the number of monkeys and the number of peaches left behind each time.
public static int get (int monkeys,int losesimplenum) {
int minnum = Monkeys+losesimplenum;
Return Getdiy (monkeys, minnum, 0, losesimplenum, monkeys);
}

public static int Getdiy (int times, int sum, int lastnum,int losesimplenum,int monkeys) {
int minnum = Monkeys+losesimplenum;
int nextmonkeys = monkeys-1;
/* SYSTEM.OUT.PRINTLN (times+ "sum:" +sum+ "\nlastnum:" +lastnum);
0 is returned
if (times = = 0) {
return sum;
}else if (times = = Monkeys) {
Start the calculation again to determine if you want to reset the data
if ((sum-losesimplenum)%monkeys==0&&sum>=minnum) {
System.out.println ("Replare");
Return Getdiy (Times-1, Sum, (sum-losesimplenum)/monkeys*nextmonkeys,losesimplenum,monkeys);
}else {
System.out.println ("replare222");
Return Getdiy (Times, Minnum, 0,losesimplenum,monkeys);
}
}else {
Calculate whether to satisfy this one time peach
if ((lastnum-losesimplenum)%monkeys==0&&lastnum>minnum) {
System.out.println ("nest");
Return Getdiy (Times-1, Sum, (lastnum-losesimplenum)/monkeys*nextmonkeys,losesimplenum,monkeys);
}else return Getdiy (monkeys, Sum+monkeys, 0,losesimplenum,monkeys);
}
}
}