Kotlin algorithm for monkeys to get started

/* Original question monkey divided peach: There are a bunch of peaches on the beach, there are five monkeys to divide. The first monkey divided the peaches into five parts, one more, and the monkey threw one more into the sea and took a copy. The second monkey to the rest of the peach is divided into five parts, and one more, it also threw a lot of one into the sea, took a copy, the third, the fifth monkey is doing this, asked: the beach originally at least how many peaches to expand the problem of monkeys: there are a bunch of peaches on the beach, there are n monkeys to divide. The first monkey divided the pile of peaches into n parts, more K, the monkey threw a lot of one into the sea, took a copy. The second monkey to the rest of the peach is divided into n parts, and more K, it also put a lot of k thrown into the sea, took a copy, until the last monkey is also asked: the beach originally at least how many peaches first traditional problem first of all, the last remaining minimum should be 6, The lowest number of peaches is obtained by reverse recursion at the base of 6. The minimum number of peaches is added when the condition is not met. The minimum number of solutions to expand the problem is n+k, so that the minimum peach count should be returned in reverse order and the number of peaches will be added n * * When the condition is not met. PackageKotlin

classMonkeypeach {//traditional monkey divided Peachoperator FunGet (Times:int, Sum:int, lastsum:int): Int {//0 returnsreturn if(Times = = 0) {sum}else if(Times = = 5) {//Restart calculation to determine if you want to reset the dataif((sum-1)% 5 = = 0 && sum >= 6) {Get (Times-1, Sum, (sum-1)/5 * 4)}Else{Get (5, 6, 0)}}Else{//calculates if the peach is satisfiedif((lastSum-1)% 5 = = 0 && lastsum > 6) {Get (Times-1, Sum, (lastSum-1)/5 * 4)}Else Get (5, sum + 5, 0)}//custom-defined monkeys include the number of monkeys and the number of the same peaches left over each timeoperator FunGet (Monkeys:int, Losesimplenum:int): Int {ValMinnum = monkeys + losesimplenumreturnGetdiy (monkeys, minnum, 0, losesimplenum, monkeys)} FunGetdiy (Times:int, Sum:int, Lastnum:int, Losesimplenum:int, monkeys:int): Int {ValMinnum = monkeys + losesimplenumValNextmonkeys = monkeys-1/* SYSTEM.OUT.PRINTLN (times+ "sum:" +sum+ "\nlastnum:" +lastnum); *///0 returnreturn if(Times = = 0) {sum}else if(Times = = monkeys) {//Restart calculation to determine if you want to reset the dataif((sum-losesimplenum)% Monkeys = = 0 && sum >= minnum)
                {//system.out.println ("Replare"); Getdiy (Times-1, Sum, (sum-losesimplenum)/monkeys * Nextmonkeys, losesimplenum, monkeys)}Else{//system.out.println ("replare222"); Getdiy (Times, Minnum, 0, losesimplenum, monkeys)}}Else{//calculates if the peach is satisfiedif((lastnum-losesimplenum)% Monkeys = = 0 && lastnum > Minnum)
                {//system.out.println ("nest"); Getdiy (Times-1, Sum, (lastnum-losesimplenum)/monkeys * Nextmonkeys, losesimplenum, monkeys)}Else Getdiy (monkeys, sum + monkeys, 0, losesimplenum, monkeys)}}