The story of a cow

There is a cow, which has a small cow at the beginning of each year.
Each heifer starts its fourth year with a heifer.
Please program implementation how many cows are there in the first n years?

/*
There is a cow, which has a small cow at the beginning of each year.
Each heifer starts its fourth year with a heifer.
Please program implementation how many cows are there in the first n years?

Title Requirements:
Input
The input data consists of multiple test instances, one for each test instance, including an integer n (0< n<), and the meaning of n as described in the topic.
N=0 indicates the end of the input data and does not handle it.
Output
For each test instance, output the number of cows in the nth year.
Each output occupies one row.
Sample Input
2
4
5
0
Sample Output
2
4

Analysis: This question is similar to another interesting classical math problem: there is a pair of rabbits, from the 3rd month after birth, every month has a pair of rabbits.
The rabbit grew to the 3rd month, and every month he gave birth to a pair of little rabbits. Suppose all the rabbits do not die, ask the total number of rabbits each month?
This is a typical Fibonacci sequence problem:
F1=1 (N=1)
F2=1 (n=2)
F3=F1+F2 (n>=3)

The number of cows is also a regular form of the sequence, so we must find a recursive formula, after analysis
F1=1 (N=1)
f2=2 (n=2)
F3=3 (n>=3)
F4=F1+F3 (n>=4)

*/

#include <iostream>using namespacestd;intMain () {intn,i,f1=1, f2=2, f3=3;//F1,F2,F3 represents the number of cows for three consecutive years, and N represents the first few yearsCin>>N; while(n!=0) {  if(n==1|| n==2|| n==3)//If the year entered is 1th, 2, 3 years{cout<<n<<Endl; CIN>>N; Continue; } F1=1; F2=2; F3=3;  for(i=1; i<= (n4)/3+1; i++)//recursive from the 4th year onwards{F1=f1+f3; F2=f1+F2; F3=f2+f3; }  if(n%3==1) cout<<f1<<Endl; if(n%3==2) cout<<f2<<Endl; if(n%3==0) cout<<f3<<Endl; CIN>>N;} return 0;} 

The story of a cow