hdu 5366 The mook jong 【dp】

The mook jong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1065    Accepted Submission(s): 702

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

 

题意：有n块砖，在砖上放东西，使东西间的间隔小于等于两块砖，问有多少种放法，最少放一样东西。

思路：放第i块砖有三种放法：
第一种：放东西且和上一个砖间隔2块。不论第i-3块砖放不放东西，这种情况都继承i-3块砖的方法总数。
第二种：不放东西。这种情况继承i-1块砖的方法总数。
第三种：放东西且前面i-1块砖都不放东西。这种情况方法+1。 

 

#include<stdio.h>

int main()
{
    int n,i;
    long long dp[65];//输出60时会超int范围  
    dp[0] = 0;
    dp[1] = 1;
    dp[2] = 2;
    for(i = 3; i <= 60; i ++)
        dp[i] = dp[i-1] + dp[i-3]+1;//状态转移方程 
    while(scanf("%d",&n)!=EOF)
    {
        printf("%lld\n",dp[n]);
    }
    return 0;
}