bzoj 2038 [2009国家集训队]小Z的袜子(hose)

\[ans_{i}=\frac{\sum_{i=l}^r \dbinom{2}{cnt_{i}}}{\dbinom{2}{r-l+1}}\]
\[ans_{i}=\frac{\sum_{i=l}^r \frac{cnt_{i}\times (cnt_{i}-1)}{2}}{\frac{(r-l+1)\times (r-l)}{2}}\]
\[ans_{i}=\frac{\sum_{i=l}^r cnt_{i}\times (cnt_{i}-1)}{(r-l+1)\times (r-l)}\]
\[ans_{i}=\frac{\sum_{i=l}^r cnt_{i}^2-cnt_{i}}{(r-l+1)\times (r-l)}\]
\[ans_{i}=\frac{\sum_{i=l}^r cnt_{i}^2 - \sum_{i=l}^r cnt_{i}}{(r-l+1)\times (r-l)}\]
因为
\[\sum_{i=l}^r cnt_{i}=r-l+1\]
所以
\[ans_{i}=\frac{\sum_{i=l}^r cnt_{i}^2 -(r-l+1)}{(r-l+1)\times (r-l)}\]

转载于:https://www.cnblogs.com/hzf29721/p/9491720.html