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