Friday, December 2, 2016

HDU 5366 - The mook jong


http://acm.hdu.edu.cn/showproblem.php?pid=5366

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
http://www.lai18.com/content/4329060.html
在1*n的院子里铺有1*1的地板,在地板上里放置木人桩,木人桩之间至少相隔2快地板,问至少放置一个木人桩的情况下,可能的情况

http://www.cnblogs.com/fenice/p/5321679.html

方法一:

  区间dp。
  dp[i][j]表示前i个地砖放j个木人桩(其中第i个地砖肯定有放东西,要不然就不会是dp[i][j],而是dp[k][j](k<i) )
  则易得转移方程 dp[i][j]=sum(dp[k][j-1]),其中k<i-2;
  初始化要注意:dp[i][1]=1,而其他则都清零。

http://blog.csdn.net/david_jett/article/details/47376251
  状态转移方程: dp[i]=dp[i-1]+dp[i-3]+1。
  dp[i]的含义是到i这个位置为止,有多少种方案数,也就是答案。因为dp表示的是合法的解,所以之前一定已经至少放了一个木桩了。dp[i-1]代表的是当前位置i不放木桩,
dp[i-3]代表的是当前位置放,因为间隔为2,所以不论前面第三个位置有没有,当前位置i都可以放置1个木桩,至于最后加上的一个1,开始没怎么理解,其实它代表的是前面i-1
个位置都没放置木桩,而在当前位置i放置一个木桩,这也是一组合法的解,故加上1。
  1. long long dp[65]={0,1,2,3};  
  2. void init()  
  3. {  
  4.   for(int i=4;i<=60;i++)  
  5.       dp[i]=dp[i-1]+dp[i-3]+1;  
  6. }  
  7. int main()  
  8. {  
  9.     init();  
  10.     int n;  
  11.     while(~scanf("%d",&n))  
  12.     {  
  13.         printf("%lld\n",dp[n]);  
  14.     }  
  15.     return 0;  
  16. }  
http://www.it610.com/article/5474922.htm
令f[i]为最后一个木人桩摆放在i位置的方案,令s[i]为f[i]的前缀和,即s[i]=f[0]+f[1]+...+f[i]。很容易就能想到f[i]=s[i-3]+1,因为最后一个木人放到最后一个位置,因此只有i-3个木人是可用的,再加上最后一种情况,s[i]=s[i-1]+f[i],而s[n]即是所求答案。本题唯一一个值得注意的点就是当n接近60时会爆int。
int main()
{
    int n,i;
    long long s[70],f[70];//s[i]表示i个木桩时的结果,f[i]表示把最后一个木桩放到第i个位置的方案数
    while(cin>>n)
    {
        s[1]=1;
        s[2]=2;
        s[3]=3;
        f[1]=f[2]=f[3]=1;
        for(i=4;i<=n;i++)
        {
            f[i]=s[i-3]+1;
            s[i]=s[i-1]+f[i];
        }
        cout<<s[n]<<endl;
    }
 
 
    return 0;
}



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