CF_Edu

Edu181-D. Segments Covering

dp[i]dp[i]: 前ii 段被精准覆盖一次的概率
dp[i]={l,r==i,p,q}dp[l1]×pq×TiTl1×qpqdp[i]=\sum_{\{l,r==i,p,q\}} dp[l-1]\times\frac{p}{q}\times\frac{T_i}{T_{l-1}\times\frac{q-p}{q}}
T[i]T[i]: 右端点在 ii 左端的线段都不存在的概率
TiTl1\frac{T_i}{T_{l-1}}: 右端点在 l,rl,r 的线段都不存在的概率

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