比赛时会错题意+不知道怎么线段树维护分数- - 思路来自题解 /* HDU 6070 - Dirt Ratio [ 二分,线段树 ] | 2017 Multi-University Training Contest 4 题意: 给出 a[N]; 设 size(l,r)为区间(l,r)不同数字的个数,求 size(l,r)/(r-l+1) 的最小值 限制: N <= 6e5, a[i] <= 6e5 分析: 二分答案 mid 则判定条件为是否存在 size(l,r)/(r-l+1) <=…

http://acm.hdu.edu.cn/showproblem.php?pid=6070 题意: 找出一个区间,使得(区间内不同数的个数/区间长度)的值最小,并输出该值. 思路: 因为是要求$\frac{f(x)}{g(x)}$的最值,所以这是分数规划的题目,对于分数规划,是要用二分查找的方式去解决的. 就像官方题解说的,二分查找mid,二分答案mid,检验是否存在一个区间满足$\frac{size(l,r)}{(r-l+1)}<=mid$,表示l~r内不同数的个数. 先把上面的式子转化一下…

题目链接 Problem Description In ACM/ICPC contest, the ''Dirt Ratio'' of a team is calculated in the following way. First let's ignore all the problems the team didn't pass, assume the team passed X problems during the contest, and submitted Y times for t…

Dirt Ratio Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1473    Accepted Submission(s): 683Special Judge Problem Description In ACM/ICPC contest, the ''Dirt Ratio'' of a team is calculated…

Dirt Ratio Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Special Judge Problem Description In ACM/ICPC contest, the ''Dirt Ratio'' of a team is calculated in the following way. First let's ignore all the proble…

目录 目录 思路: (有任何问题欢迎留言或私聊 && 欢迎交流讨论哦 目录 题意:传送门  原题目描述在最下面.  求\(sum/len\)最小值.\(sum\)是一段区间内不同数字的个数,\(len\)是这段区间的长度. 思路:  首先预处理出每个数上一次出现的位置\(pre[i]\)和最后一次出现的位置\(lst[i]\).这个操作在静态求区间内不同数的个数和动态求区间内不同数的个数都有用到. 法一:  二分答案\(mid\).枚举序列,每加入一个数就在\(pre[i]-i\)区间加一…

题 OvO http://acm.hdu.edu.cn/showproblem.php?pid=6070 (2017 Multi-University Training Contest - Team 4 - 1004) 解 二分答案 check时,要满足distinct(l,r)/(r-l+1)<val ,将这个不等式转化为distinct(l,r)+val*l<val*(r+1) check的时候,从左到右枚举右端点r,用线段树维护查询从1到r中选一个l,distinct(l,r)+val*…

Dylans loves tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1444    Accepted Submission(s): 329 Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes…

Multiply game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3224    Accepted Submission(s): 1173 Problem Description Tired of playing computer games, alpc23 is planning to play a game on numbe…

HDU 1394 Minimum Inversion Number(线段树求最小逆序数对) ACM 题目地址:HDU 1394 Minimum Inversion Number 题意:  给一个序列由[1,N]构成.能够通过旋转把第一个移动到最后一个.  问旋转后最小的逆序数对. 分析:  注意,序列是由[1,N]构成的,我们模拟下旋转,总的逆序数对会有规律的变化.  求出初始的逆序数对再循环一遍即可了. 至于求逆序数对,我曾经用归并排序解过这道题:点这里.  只是因为数据范围是5000.所以全…