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…

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…

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*…

区间交 Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 849    Accepted Submission(s): 377 Problem Description 小A有一个含有n个非负整数的数列与m个区间.每个区间可以表示为li,ri. 它想选择其中k个区间, 使得这些区间的交的那些位置所对应的数的和最大. 例如样例中,选择[2,5]…

Flipping Parentheses 题目连接: http://codeforces.com/gym/100803/attachments Description A string consisting only of parentheses '(' and ')' is called balanced if it is one of the following. • A string "()" is balanced. • Concatenation of two balance…

Intervals 题目连接: http://codeforces.com/gym/100231/attachments Description 给你n个区间,告诉你每个区间内都有ci个数 然后你需要找一个最小的点集,使得满足这n个区间的条件 Input n 然后li,ri,ci Output 输出最小点集大小 Sample Input 5 3 7 3 8 10 3 6 8 1 1 3 1 10 11 1 Sample Output 6 Hint 题意 题解: 线段树+二分+贪心 首先我们贪心一…

枚举左端点,然后在线段树内,更新所有左边界小于当前点的区间的右端点,然后查线段树二分查第k大就好 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long LL; ; LL a[N]; struct Node{ int l,r; bool operator<(const Node &rhs)const{ if(l==rhs…