算法----bonus dumplings - L.J.SHOU的专栏 - 博客频道 - CSDN.NET


算法----bonus dumplings - L.J.SHOU的专栏 - 博客频道 - CSDN.NET
过年了,妈妈做了100只饺子,其中有10只饺子里面有1块的硬币。
小明依次吃这100只饺子,如果小明连续吃到k个硬币,那么小明得到k-1个硬币。
e.g. 110111表示6只饺子,1表示有硬币,0表示没有。11表示连续吃到2个饺子,那么小明得1个硬币;111连续迟到3个,小明得2个硬币;故,小明共得到3个硬币。
问小明得到的硬币的期望值是多少?

分析

期望定义: E(X)=iXiP(Xi)
在本题中,随机事件X即为小明最终得到的硬币数目,x[0,9]
  • 计算P(Xi)=cases(x=i)allcases
  • 计算期望
那么原问题就简化为小明得到硬币k,所有可能的cases的数目。
采用动态规划,子问题定义如下
  • f[i][j][k]――前i个饺子,含j个有硬币的饺子,得分是k,且第i个饺子不含硬币,所有可能的情况的总数。
  • g[i][j][k]――前i个饺子,含j个有硬币的饺子,得分是k,且第i个饺子含硬币,所有可能的情况的总数。
那么递推公式如下,
  • f[i][j][k]―-第i个饺子不包含硬币,所以不用第i个饺子
    • f[i][j][k] = f[i-1][j][k] + g[i-1][j][k]
  • g[i][j][k]―-第i个饺子包含硬币,那么这个硬币能够得到,取决于第i-1个饺子是否包含硬币
    • g[i][j][k] = f[i-1][j-1][k] + g[i-1][j-1][k-1]

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