CF653F Paper task】的更多相关文章

[CF653F] Paper task Description 给定一个括号序列,统计合法的本质不同子串的个数. Solution 很容易想到,只要在传统统计本质不同子串的基础上修改一下即可. 考虑经典统计过程,对于第 \(i\) 个后缀,它的贡献为 \(n - sa[i] + 1 - h[i]\) 也就意味着,它产生贡献的区间是 \([sa[i]+h[i], n]\) .换言之,对任意 \(j \in [sa[i]+h[i], n]\) , \(s[sa[i],j]\) 是一个答案. 那么我们…
题目链接:洛谷 首先我们不考虑本质不同这个限制. 既然不能直接用栈乱搞,我们就可以用一个前缀和的套路了. 我们将(设为1,将)设为-1,记前缀和为$s_i$,则$[i,j]$这一段是回文子串当且仅当 1.$s_j=s_{i-1}$ 2.$\forall k\in [i,j],s_k\geq s_{i-1}$ 于是我们枚举$i$,显然$j$要满足第二个性质就肯定不能超过一个上界,这个上界是可以二分的.每次check的时候就判断一下区间最小值,可以用ST表维护. 然后看看本质不同如何做. 这时候我们…
Paper task 如果不要求本质不同直接st表二分找出最右端, 然后计数就好了. 要求本质不同, 先求个sa, 然后用lcp求本质不同就好啦. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII…
题目链接:http://codeforces.com/problemset/problem/653/F 大意是给出一个只包含'('和')'的括号串,求有多少不同的子串是合法的括号串 解法:对于每一个后缀,需要能够求出这个后缀有多少前缀是合法的括号串,这个可以用O(log n)复杂度的二分来解决.注意,二分的范围并不是整个后缀,因为如果将'('视作+1, ')'视作-1,则一个合法的括号串必须时刻不能小于0.所以可以在ST表上二分出合法范围,在这个范围内去统计有多少合法串(即'('与')'正负相消…
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