[ZJOI2017]仙人掌

对于链:
(f[i])为前(i)点的方案数
(f[i]=sum_{k=1}^{i-1}f[k])
(f[i+1]=sum_{k=1}^{i}f[k])=>(f[i+1]=2*f[i])

戒骄戒躁
原文地址:https://www.cnblogs.com/lxzl/p/10433704.html