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

#include<iostream>
#include<malloc.h>
#include<cstdio>
using namespace std;
int main()
{
    int t,a,b,n,*sum,i,j;
    while(scanf("%d",&t)!=EOF&&t<=10)
    {
      while(t--)
      {
        scanf("%d%d%d",&a,&b,&n);
        sum=(int *)malloc(sizeof(int)*(n+1));
        sum[1]=a;sum[2]=b;
        for(i=3;i<=n;i++)
        {//判断下标是偶还是奇数
            if(i%2!=0) sum[i]=sum[i-1]+sum[i-2];
            else sum[i]=sum[i-1]+sum[i-2]+sum[i-3];
        }
        printf("%d\n",sum[n]);
        free(sum);
       }
    }
    return 0;
}

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