Codeforces 590 A:Median Smoothing】的更多相关文章

A. Median Smoothing time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article…
C. Median Smoothing Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/problem/C Description A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that des…
B. Rebranding The name of one small but proud corporation consists of n lowercase English letters. The Corporation has decided to try rebranding — an active marketing strategy, that includes a set of measures to change either the brand (both for the…
题目链接: A. Median Smoothing time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia a…
C. Median Smoothing   A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and e…
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the meth…
A. Median Smoothing time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article…
题目链接:http://codeforces.com/problemset/problem/590/A 题目大意是给一个串,头和尾每次变换保持不变. 中间的a[i]变成a[i-1],a[i],a[i+1]的中位数,而且此题串是01串. 对于01串 0 0 0中位数是0 0 0 1中位数是0 0 1 1中位数是1 1 1 1中位数是1 所以 1.串中有两个相邻以上的0或者1是保持不变的. 2.会变的只有是两个1中间的0或者两个0中间的1. 但是到这里的话,虽然证明了肯定能变成稳定态,就算把相同数字…
http://codeforces.com/problemset/problem/590/A: 在CF时没做出来,当时直接模拟,然后就超时喽. 题意是给你一个0 1串然后首位和末位固定不变,从第二项开始到倒数第二项,当前的a[i]=(a[i-1],a[i],a[i+1])三项排序后的中间项,比如连续3项为 1 0 1,那么中间的就变为1,然后题目让你输出达到稳定状态时所需的最小步数,不能的话输出-1. 无论给你啥数列,都能达到稳态.所以不可能输出-1: 还有一开始就稳定不变,或经过几次变换而稳定…
You are given an integer sequence a1,a2,…,ana1,a2,…,an. Find the number of pairs of indices (l,r)(l,r) (1≤l≤r≤n1≤l≤r≤n) such that the value of median of al,al+1,…,aral,al+1,…,ar is exactly the given number mm. The median of a sequence is the value of…
构造题. 答案可以o(n)构造出来.首先要发现规律.只有01交替的串才可能变化,变化规律如下: 1开头,长度为偶数(0结尾):变(len-2)/2次 变完后 前半1 后半01开头,长度为奇数(1结尾):变(len-1)/2次 变完后 全为10开头,长度为偶数(1结尾):变(len-2)/2次 变完后 前半0 后半10开头,长度为奇数(0结尾):变(len-1)/2次 变完后 全为0 然后就是在原串中寻找01交替串,然后按照上述规律进行变换. #include <cstdio> #include…
题意:一个数组,一次操作为:除首尾不变,其它的=它与前后数字的中位数,这样对数组重复几次后数组会稳定不变.问要操作几次,及最后的稳定数组. 挺难的题,参考了别人的代码和思路.总的来说就是找01010,010101这样的,分别变为00000,000111.不知道谁发现的将它和异或联系起来.比如对010101异或前一个元素为011111(记为数组b),数组b中连续的1全置0,如果是奇数那么连续的1的中间置1.结果是000100(数组c).再对数组c(除第一个元素)异或前一个元素,得到000111就是…
分析: 三个01组合只有八种情况: 000 s001 s010 0011 s100 s101 1110 s111 s 可以看出只有010,101是不稳定的.其他都是稳定的,且连续地出现了1或0,标记为s. 考虑连续的不稳定串的,例子: 11010100   s        s  110100    s    s    1100 只有两种情况,两个边界是不同(11和00)或者相同(11或者00). 前者中间的不稳定串的长度是2*k,且经过k次将变成11110000,而后者的不稳定串的长度是2*k…
问题: 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)). Example1:nums1 = [1, 3]nums2 = [2]The median is 2.0Example2:nums1 = [1, 2]n…
题目: There are two sorted arrays A and B 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)). 解题思路: 这道题,很容易想到的是先排序再直接定位中间那个数即可,m+n为偶数的话,应为中间两数之和除2,但时间复杂度不符合题目要求,还一种方法就是利用归并思想,因…
http://codeforces.com/gym/101142/attachments 题意:每个人在TC和CF上分别有两个排名,如果有一个人在任意一个网站上大于另一个人的排名,那么这个人可以打败另外一个人.还有就是如果 B 能打败 A, C 能打败 B,但是 C 直接从排名上看 C 并不能打败 A,但是因为 B -> A 并且 C -> B,所以 C -> B -> A, 即 C 也能(通过打败 B 来)打败 A. 如这个样例: A :5, 5 , B :1, 6,  C:2,…
Median of Two Sorted Arrays There are two sorted arrays A and B 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)). time=378ms accepted <pre name="code" class="j…
题目难度:hard 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)). Example 1: nums1 = [1, 3] nums2 = [2] The median is 2.0 Example 2: n…
今天要分享的这篇论文是我个人最喜欢的论文之一,它的思想简单.巧妙,而且效果还相当不错.这篇论文借助数学上的 \(L_0\) 范数工具对图像进行平滑,同时保留重要的边缘特征,可以实现类似水彩画的效果(见下图). 另外这篇论文的作者徐立也是一个相当高产的研究员. 论文的目的 所谓图像平滑,就是突出图像中的低频成分,抑制高频成分,减小突变的梯度.大部分情况下,这么做的目的是为了去除图像中的噪声,因为噪声一般就是一些孤立的像素点,是像素变化比较大的区域.在传统的图像处理中,大部分操作都是用一些具有平滑性…
题目链接:http://ac.jobdu.com/problem.php?pid=1004 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: // // 1004 Median.cpp // Jobdu // // Created by PengFei_Zheng on 26/04/2017. // Copyright © 2017 PengFei_Zheng. All rights reserved. // #include <std…
[Q4] 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)). You may assume nums1 and nums2 cannot be both empty. Example 1: nums1 = […
here 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)). Example 1: nums1 = [1, 3] nums2 = [2] The median is 2.0 Example 2: nums1 = [1,…
题目难度:hard 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)). Example 1: nums1 = [1, 3] nums2 = [2] The median is 2.0 Example 2: n…
D. Array Restoration 题目链接:https://codeforces.com/contest/1023/problem/D 题意: 给出一个序列,现在要求对一个全为0的序列执行q次操作,每次操作都要选定一段区间然后将区间上面的值变为i(i为操作的次数).最终使得0序列变为之前给出的序列. 原序列中如果存在0,那么说明这个值是任意的. 最后要求输出经过q次操作之后的序列. 题解: 我们首先可以想到不可行的情况:在原序列中若存在一个数a,假设其出现次数大于1,那么两端为a的中间区…
题目链接:http://codeforces.com/problemset/problem/571/B 题意: 给你一个长度为n的数列a[i]. 现在你可以随意改变数字的位置,问你 ∑| a[i] - a[i+k] | 的最小值(1 <= i <= n-k). 题解: 将a[i]拆成若干个子序列s[j],子序列中相邻两数在a[i]中的距离为k. 此时原式 = ∑(子序列s[j]内部之差的和) 显然,要想使子序列s[j]内部之差的和尽可能小,子序列s[j]内部一定为升序. 显然,要想使 ∑(子序…
题目链接:http://codeforces.com/problemset/problem/455/C 题意: 给你一个森林,n个点,m条边. 然后有t个操作.共有两种操作: (1)1 x: 输出节点x所在树的直径. (2)2 x y: 如果x,y不在同一棵树上的话,用一条边连接x,y所在的树,并且使得到的新树的直径尽可能小. 题解: 首先对于初始状态,算出每一棵树的直径d[find(i)]. 每次合并树的时候,因为要尽可能让新树直径变小,所以显然这条边要分别连接两棵树直径的“中点”. 所以新树…
题目链接:http://codeforces.com/problemset/problem/180/C 题意: 给你一个字符串s,长度为n. 让你将这个字符串变成“前面一段都是大写字母,后面一段都是小写字母”的形式. (也可以全是大写或全是小写) 问你最少改动几个字符. 题解: 表示状态: dp[i][0/1] = min changes 表示考虑到第i个字符,s[i]改成了大写(1)或小写(0)时的最小代价 找出答案: ans = min(dp[n][0], dp[n][1]) 如何转移: d…
<?xml version="1.0"?><xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <!-- Licensed to the Apache Software Foundation (ASF) under one or more contributor license agreements. See t…
#include <stdio.h> #include <stdlib.h> #include <limits.h> #define N 1000000 int a1[N+1], a2[N+1]; int cmp(const void *a, const void *b) { return *(int *)a - *(int *)b; } int main(void) { int n1, n2, i1, i2, k, id, res; while (scanf(&quo…
Median Sum You are given N integers A1, A2, ..., AN. Consider the sums of all non-empty subsequences of A. There are 2N−1 such sums, an odd number. Let the list of these sums in non-decreasing order be S1, S2, ..., S2N−1. Find the median of this list…