Arthur and Table CodeForces - 557C 首先,按长度排序. 长度为p的桌腿有a[p]个. 要使得长度为p的桌腿为最长,那么要按照代价从小到大砍掉sum{长度不到p的腿的数量}-a[p]+1条腿.还需要将所有长于p的桌腿砍光.枚举p即可. 要点(看了题解才明白):可以通过精力最高只有200的条件,大大缩小时间/空间复杂度. 恩,所以...这是贪心吧? #include<cstdio> #include<algorithm> using namespace…

C. Arthur and Table Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/557/problem/C Description Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable. In t…

C. Arthur and Table time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstab…

C. Arthur and Table time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstab…

Arthur and Table 题意 一个桌子有n个腿,每个腿都有一个高度,当且仅当最高的腿的数量大于桌子腿数量的一半时,桌子才是稳定的.特殊的是当只有一个腿时,桌子是稳定的,当有两个腿时两个腿必须都得是最高的,才稳定. 分析 这题其实和去年的牛客的一道砍树题一样的(但是我没想起来那题,当时看的题解,稍微改了一下代码,就交了印象不深刻...) Governing sand 首先我们按照桌子腿的高度从小到大排序,然后枚举当前最高的高度. 大于当前高度的肯定都要删除掉,用一个suf后缀来维护. 我…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable. In total the table A…

题意:一个桌子有n条腿,每条腿有一定的长度l,和砍下的花费w,现在规定,桌子稳的条件是长度最长的腿(可多个)的数量大于长度小于它的桌子腿数量,且不存在比他还长的桌子腿,求让桌子腿稳定的最小的花费 #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include &…

Bizon the Champion isn't just charming, he also is very smart. While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the i…

大意: 给定格点图, 每个'.'的连通块会扩散为矩形, 求最后图案. 一开始想得是直接并查集合并然后差分, 但实际上是不对的, 这个数据就可以hack掉. 3 3 **. .** ... 正解是bfs, 一个点被扩散当且仅当它所在的某个2*2块中只有它为'*'. #include <iostream> #include <iostream> #include <algorithm> #include <cstdio> #include <math.h&…

大意: 给定序列$a$, 某些位置为'?', 求给'?'赋值使得序列$(a_1+a_2+...+a_k,a_2+a_3+...+a_{k+1},...)严格递增, 且$\sum|a_i|$最小. 化简一下可以得到$a_1<a_{k+1}<a_{2k+1}<...,a_2<a_{k+2}<a_{2k+2}<...$, 所以每一部分都是独立的, 所以单独考虑k个部分, 贪心使得$\sum|a_i|$最小即可. #include <iostream> #inclu…