The mook jong
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 226 Accepted Submission(s): 167
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).
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
题目大意：在由1*1的地砖铺成的1*n的院子里摆放木桩，每个木桩放在一个地板上。每两个木桩之间间隔至少两个地板，问有多少种放法。
解题思路：比赛的时候用的搜索，那叫一个慢，就打了40——60之间的表，算是蒙混过关。但是其实正解是dp。对于某个地板，可以放或者不放，那么如果放的话，结果加上dp[i-3]。如果不放的话，dp[i-1]是最优子结构，同时加上一共只放一个木桩在i位置的情况。dp[i]=dp[i-3]+dp[i-1]+1。结果超int需注意。
#include<bits/stdc++.h>
using namespace std;
typedef long long INT;
INT dp[65];
int main(){
dp[1]=1,dp[2]=2,dp[3]=3;
for(int i=4;i<=62;i++)
dp[i]=dp[i-3]+dp[i-1]+1;
int n;
while(scanf("%d",&n)!=EOF){
printf("%lld
",dp[n]);
}
return 0;
}