SZ斐波拉契数列

Problem Description

Fibonacci sequence is familiar to you, right? Here comes the SZ Fibonacci sequence, with the property like the following formula.
F(n)= a (n=1)
F(n)= b (n=2)
F(n)= F(n-1)+F(n-2), n>2 and n is odd
F(n)= F(n-1)+F(n-2)+F(n-3), n>2 and n is even
Here a and b are constants. Given a,b,and n, your task is to calculate the F(n).

Input

First line of input comes a positive integer T(T<=10), indicating the number of test cases. Each test case contains three positive integer a,b and n(a<=10,b<=10,n<=30)

Output

Print one line containing an integer, i.e.F(n), for each test case.

Sample Input

2
1 2 3
1 3 6

Sample Output

3
24

Author

HYNU 

题意分析：

按规律些代码。

代码：

#include<stdio.h>
int fun(int a,int b,int n)//四个公式
{
   if(n==1)return a;
   else if(n==2)return b;
   if(n%2==1)return fun(a,b,n-1)+fun(a,b,n-2);
   else return fun(a,b,n-1)+fun(a,b,n-2)+fun(a,b,n-3);
}
int main()
{
   int a,b,n;
   int t;
   scanf("%d",&t);
   while(t--)
   while(scanf("%d%d%d",&a,&b,&n)!=EOF)
   {
      printf("%d\n",fun(a,b,n));
   }
   return 0;
}