HDU - 5366 The mook jong (dp动态规划)

The mook jong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 62    Accepted Submission(s): 40

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
The mook jong
 
 Accepts: 506
 
 Submissions: 1281
 Time Limit: 2000/1000 MS (Java/Others)
 
 Memory Limit: 65536/65536 K (Java/Others)
问题描述
ZJiaQ为了强身健体，决定通过木人桩练习武术。ZJiaQ希望把木人桩摆在自家的那个由1*1的地砖铺成的1*n的院子里。由于ZJiaQ是个强迫症，所以他要把一个木人桩正好摆在一个地砖上，由于木人桩手比较长，所以两个木人桩之间地砖必须大于等于两个，现在ZJiaQ想知道在至少摆放一个木人桩的情况下，有多少种摆法。
输入描述
输入有多组数据，每组数据第一行为一个整数n(1 < = n < = 60)
输出描述
对于每组数据输出一行表示摆放方案数
输入样例
123456
输出样例
1235812
/*Author: 2486Memory: 1404 KBTime: 0 MSLanguage: C++Result: Accepted*/#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;typedef long long LL;const int MAXN = 60 + 5;LL dp[MAXN][2];int n;void init() {    dp[1][0] = 0,dp[1][1] = 1;    dp[2][0] = 1,dp[2][1] = 1;    dp[3][0] = 2,dp[3][1] = 1;    for(int i = 4; i < MAXN; i ++) {        dp[i][0] = dp[i - 1][0] + dp[i - 1][1];//如果我此时这个位置不放桩子,那么状态可以从dp[i - 1][0] 以及dp[i - 1][1]传递        dp[i][1] = dp[i - 3][0] + dp[i - 3][1] + 1;//如果放桩子,可以从前两个桩子前传递过来状态    }}int main() {    init();    //freopen("D://imput.txt", "r", stdin);    while(~ scanf("%d", &n)) {        printf("%I64d\n", dp[n][0] + dp[n][1]);    }    return 0;}