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  • 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
    ![](../../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
    ",dp[n]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/hellocheng/p/7521334.html
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