HDU 6070 线段树】的更多相关文章

HDU 6070 Partition Problem : 给一段长度为n的序列,要求找出一段区间,使得这段区间的数字种类除以区间长度最小.输出最后的答案即可.(n <= 60000)(9s时限) Solution : 显然,答案是0~1中的一个数字,可以很自然的想到二分答案的做法.假设目前二分到的答案为mid,那么需要判断 \[\frac{cnt(l, r)}{r- l + 1} <= mid\] 其中cnt(l,r)为l到r这个区间内的数字种类.变化一下式子可以得到: \[cnt(l, r)…
题意:求AC率,x/y 的最小值,x是区间数字的种类数,y是区间的长度. 分析: 二分答案比率.ans, 动态插入结点,一些区间的size会发生变化,是那些前面暂时没有新的结点的区间 size + 1. ans*l,每一个区间只有一个ans*l,只与 l 相关,线段树单点更新. 用线段树维护区间的最小值.最小值小于 ans,二分左移. #include <bits/stdc++.h> using namespace std; ; ; int a[maxn]; int last[maxn]; ]…
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