用递归解决猴子吃桃问题

#include<bits/stdc++.h>

using namespace std;

int num = 6;//桃子个数 

int f(int n,int b){
    if(n==0) return 1;
    if((b-1)%5 !=0){
    	return 0;
	}
	b = (b-1)*4/5;

	return f(n-1,b);
		
}
int main(){
	while(1){
		num+=1;
		bool flag = f(5,num);
		//cout<<flag<<endl;
		if(flag){
			cout<<num<<endl;
			break;
		}
	}
}