HDU 5366 The mook jong(递推)

The mook jong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 365    Accepted Submission(s): 283

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

123456

 

Sample Output

1235812
1*n的格子 ，每个格子可以放一个木人桩，相邻的木人桩最少隔 两个格子 ，求至少放一个木人桩的不同放法数
 f[i] = f[i-1] + f[i-3] + 1
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <stdlib.h>#include <cmath>#include <map>#include <queue>#define inf 0x3f3f3f3fusing namespace std;int main(){        __int64 b[65];        b[1]=1;        b[2]=2;        b[3]=3;        for(int i=4;i<=60;i++)        {            b[i]=b[i-1]+b[i-3]+1;        }        int n;    while(~scanf("%d",&n))    {        printf("%I64d\n",b[n]);    }    return 0;}