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  • HDU5366——The mook jong——dp

    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
    ![](../../data/images/C613-1001-1.jpg)

    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;
    }
    

      

     
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  • 原文地址:https://www.cnblogs.com/chengsheng/p/4714839.html
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