https://www.luogu.org/problem/show?pid=2115 题目描述 Farmer John's arch-nemesis, Farmer Paul, has decided to sabotage Farmer John's milking equipment! The milking equipment consists of a row of N (3 <= N <= 100,000) milking machines, where the ith machi…

题目描述 Farmer John's arch-nemesis, Farmer Paul, has decided to sabotage Farmer John's milking equipment! The milking equipment consists of a row of N (3 <= N <= 100,000) milking machines, where the ith machine produces M_i units of milk (1 <= M_i &…

题目描述 Farmer John's arch-nemesis, Farmer Paul, has decided to sabotage Farmer John's milking equipment! The milking equipment consists of a row of N (3 <= N <= 100,000) milking machines, where the ith machine produces M_i units of milk (1 <= M_i &…

还是二分答案,发现我的$check$函数不太一样,来水一发题解 列一下式子 $$\frac{sum-sum[l,r]}{n-(r-l+1)}<=ans$$ 乘过去 $$sum-sum[l,r]<=ans*(n-r+l-1)$$ 即 $$\sum_{i=1}^{l-1}+\sum_{i=r+1}^{n}<=ans*(n-r+l-1)$$ $$\sum{(a_i-ans)}<=0$$ 所以我们在$check$函数中,可以处理出$a_i-ans$数组 然后求个前缀和,后缀和,前缀最小值,…

题意:给你一个正整数序列,让你删去一段区间内的数[l,r] $1<l\le r <n$ 使得剩余的数平均值最小$n\le 10^5$ 1.不难想到暴力,用前缀和优化$O(n^2)$ #include<cstdio> #include<iostream> #include<cstring> #include<cctype> #include<algorithm> #include<cmath> using namespace…

我对二分的理解:https://www.cnblogs.com/AKMer/p/9737477.html 题目传送门:https://www.luogu.org/problemnew/show/P2115 对于我们要求的一个"最小平均值",我们可以通过二分来得到.对于我们二分的那个平均值,我们令每一个数全部减去它,然后这时删掉"最大子段和"就是最优策略. 假设减完平均值之后的数列和为\(sum\),那么我们二分的平均值为\(ave\),要使得平均值降到\(ave\)…

本来是想找一道生成树的题做的...结果被洛咕的标签骗到了这题...结果是二分答案与生成树一点mao关系都没有.... 题目大意:给你一个序列,请你删去某一个$l~r$区间的值($2<=i<=j<=n-1$),使得剩余元素的平均值最小. 开始是想二分序列长度的,后来发现没什么卵用...于是再想一想二分平均值,但是又感觉并没有二分单调性...(其实是满足的,因为我们二分出的最终答案,当比这个答案大的时候,我们一定能满足,小的时候一定不能满足.) 因为二分的复杂度带了一个$log$,所以我们$…

题目链接 Solution 去掉中间一段区间 \([l,r]\) 后剩下的平均值可以表示为 : \[\frac{\sum^{n}_{i=1}{v_i}-\sum^{r}_{i=l}{v_i}}{n-(r-l+1)}\] 二分的答案如果要满足条件,即: \[\frac{\sum^{n}_{i=1}{v_i}-\sum^{r}_{i=l}{v_i}}{n-(r-l+1)}<Ans\] 移一下项,然后可以得到: \[{\sum^{n}_{i=1}{(v_i-Ans)}<\sum^{r}_{i=l}{…

给定一串数,问删除中间一段,剩下的平均数最小是多少: 不容易想到这是个二分. $solution:$ 来手玩一点式子: 首先很容易想到一个前缀和$sum_i $表示i到1的前缀和,这样就能很容易地O(1)查询区间和/差 二分一个mid,作为最小的平均数. 假设删去区间为l-r(lr都删) 平均数等于: $ave={sum_{i-1} + sum_{n-j} }/{n-(j-i+1)}$ 于是,这里就就有了单调性: $mid \leq ave$ 稍微变形一下,就得到 $sum_n-sum_j+su…

洛谷 P2115 [USACO14MAR]破坏Sabotage https://www.luogu.org/problem/P2115 JDOJ 2418: USACO 2014 Mar Gold 2.Sabotage https://neooj.com/oldoj/problem.php?id=2418 Description Problem 2: Sabotage [Brian Dean, 2014] Farmer John's arch-nemesis, Farmer Paul, has…