C. Maximum Median 题意: 给定一个数组,可每次可以选择一个数加1,共执行k次,问执行k次操作之后这个数组的中位数最大是多少? 题解:首先对n个数进行排序,我们只对大于中位数a[n/2]的数进行操作,所以这个最大中位数的取值范围是确定的,在区间[ [a[n/2],a[n-1] ]之内,二分枚举最大的中位数x; 通过判断使x成为最大中位数的操作数是否大于k来缩小范围 #include<iostream> #include<string.h> #include<…
Median Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12453 Accepted: 4357 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < j ≤ N). We can get C(N,2) differenc…
题目链接:http://poj.org/problem?id=3579 Median Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8286 Accepted: 2892 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < …
Median Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5118 Accepted: 1641 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < j ≤ N). We can get C(N,2)differences…
题目链接: B. Maximum Value time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given a sequence a consisting of n integers. Find the maximum possible value of (integer remainder of ai divi…
Median http://poj.org/problem?id=3579 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11225 Accepted: 4016 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < j ≤ …
Moo University - Financial Aid Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4235 Accepted: 1293 Description Bessie noted that although humans have many universities they can attend, cows have none. To remedy this problem, she and he…
Moo University - Financial Aid Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 6599 Accepted: 1926 Description Bessie noted that although humans have many universities they can attend, cows have none. To remedy this problem, she and he…
A. Important Exam 水题 #include<iostream> #include<string.h> #include<algorithm> #include<stdio.h> using namespace std; ; char a[maxx][maxx]; int pre[maxx]; int b[maxx]; int main(){ int n,m; while(~scanf("%d%d",&n,&…
Moo University - Financial Aid Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 6020 Accepted: 1792 Description Bessie noted that although humans have many universities they can attend, cows have none. To remedy this problem, she and her fe…
Data manipulation primitives in R and Python Both R and Python are incredibly good tools to manipulate your data and their integration is becoming increasingly important1. The latest tool for data manipulation in R is Dplyr2 whilst Python relies onPa…
Moo University - Financial Aid Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 7961 Accepted: 2321 Description Bessie noted that although human…
Description Bessie noted that although humans have many universities they can attend, cows have none. To remedy this problem, she and her fellow cows formed a new university called The University of Wisconsin-Farmside,"Moo U" for short. Not wish…
| Main | Site Index | Download | mimetic A free/GPL C++ MIME Library mimetic is a free/GPL Email library (MIME) written in C++ designed to be easy to use and integrate but yet fast and efficient. It is based on the C++ standard library and heavily us…
Moo University - Financial Aid Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 10894 Accepted: 3206 Description Bessie noted that although humans have many universities they can attend, cows have none. To remedy this problem, she and her f…
GV900 Political Explanation, 2017/201830 October, 2018Homework assignment 2Due Week 7 (13 November)Write an R code file named (gv900-HW2.R) to complete the following tasks. The easiestway to write an R code file is to start with an existing file: Dup…
Description Bessie noted that although humans have many universities they can attend, cows have none. To remedy this problem, she and her fellow cows formed a new university called The University of Wisconsin-Farmside,"Moo U" for short. Not wish…
比赛链接:https://codeforc.es/contest/1201 A. Important Exam 题意:有\(n\)个人,每个人给出\(m\)个答案,每个答案都有一个分值\(a_i\),每个问题的正确答案不确定,询问最大可能的得分为多少. 分析:对于每个问题贪心最大数量就好. AC代码: #include <bits/stdc++.h> #define SIZE 500007 #define rep(i, a, b) for(int i = a; i <= b; ++i)…
Delay Constrained Maximum Capacity Path Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1839 Description Consider an undirected graph with N vertices, numbered from 1 to N, and M edges. The vertex numbered with…
题目链接:Codeforces 484B Maximum Value 题目大意:给定一个序列,找到连个数ai和aj,ai%aj尽量大,而且ai≥aj 解题思路:类似于素数筛选法的方式,每次枚举aj,然后枚举k,每次用二分找到小于k∗aj而且最大的ai,维护答案,过程中加了一些剪枝. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn =…
Median Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11599 Accepted: 4112 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < j ≤ N). We can get C(N,2) differences t…
D - Median of Medians Time limit : 2sec / Memory limit : 1024MB Score : 700 pointsProblem Statement We will define the median of a sequence b of length M, as follows: Let b' be the sequence obtained by sorting b in non-decreasing order. Then, the…
There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 看到这道题的通过率很诧异,感觉这道题挺容易的,因为其实它的思想还是很简单的. 1)最笨的方法去实现,利用排序将两个数组合并成一个数组,然后返回中位数,这种方…
4. Median of Two Sorted Arrays 题目链接:https://leetcode.com/problems/median-of-two-sorted-arrays/ Description: There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time compl…
