Hdu5366 The mook jong

Problem Description

ZJiaQ want to become a strong man, so he decided to play the mook jong。ZJiaQ want to put some mook jongs in his backyard. His backyard consist of n bricks that is 1*1,so it is 1*n。ZJiaQ want to put a mook jong in a brick. because of the hands of the mook jong， the distance of two mook jongs should be equal or more than 2 bricks. Now ZJiaQ want to know how many ways can ZJiaQ put mook jongs legally(at least one mook jong).

Input

There ar multiply cases. For each case, there is a single integer n( 1 < = n < = 60)

Output

Print the ways in a single line for each case.

Sample Input

1
2
3
4
5
6

Sample Output

1
2
3
5
8
12

分别用a[i],b[i]表示总共位置数为i时，最后一个含桩和不含桩的种数，那么所求结果即为a[i]+b[i];
对于b[i]，由于其第i个位置前面情况可以任意，故b[i]=a[i-1]+b[i-1];
而对于a[i]，只需第i-2、i-1个位置为空；
同时考虑到以该方法计算时，会遗漏前面全为0的情况，故在加一即可，即a[i]=b[i-2]+1。

#include 
#include 
int main()
{
    int n,i;
    long long a[65],b[65];
    b[1]=0;
    b[2]=1;
    b[3]=2;
    a[1]=1;
    a[2]=1;
    a[3]=1;
    for(i=4;i<61;i++){
        a[i]=b[i-2]+1;
        b[i]=b[i-1]+a[i-1];
    }
    while(~scanf("%d",&n)){
        printf("%lld\n",a[n]+b[n]);
    }
    return 0;
}