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Qt6 QML Book/Qt for Python/局限性

Limitations 局限性 At the moment, there are some things that are not easily available. One of them is that you cannot easily create QML plugins using Python. Instead you need to import the Python QML “modules” into your Python program and then use qmlRegi

游戏术语

ARAM=All Random All Middle (所有人随机选择英雄,所有人都走中路)   FEEDER means Player Who Dies Easily in a Game in an online gaming context. A FEEDER is a player who repeatedly "Dies Easily in a Game" and thereby feeds the opposition with skills, experience

Perform Easily CodeForces - 1413C

原题链接 题意:求构造c[k] = b[i]-a[j],求(最大差值-最小差值)的最小值 考察:双指针 错误思路:   排序b,a.输出b[n]-a[n]-b[1]+a[1].实际上最大差值不一定由b[n]构成,最小差值不一定由b[1]构成.如果b数组都相同,最大差值和最小差值可以是同一个数. 正确思路:   算出所有b[i]-a

Tutorial: How to "Easily" Play Past Brawls/Old Maps in Single Player

Tutorial: How to "Easily" Play Past Brawls/Old Maps in Single Player Without further work that I haven't bothered to figure out yet be warned this means you're alone, no AI on either team. So something like Pull Party is pretty lame

CF1413C Perform Easily 题解

毒瘤C题,考场卡我1个小时 首先,这道题难点在哪里?它的最大值与最小值都是浮动的。 怎么办?把最小/最大值固定! 以把最小值固定为例,我们枚举每个音符,并枚举它使用哪条琴弦,将它此时的位置强制其作为最小值(设为\(minx\))。 同时,我们令其他音符不作为最小值,即其他的音符的位置不能小于最小值

codeforces C - Perform Easily (尺取法 + 线段树)

题目链接:https://codeforces.com/contest/1435/problem/C 给定 \(n\) 个 \(b_i\),每个 \(b_i\) 可以选择减去\(a_{1,\ldots,6}\)中的一个数字,求新数列中最大值减最小值的最小值 题意很绕,需要仔细理解 将所有的二元组\((b_i-j,i)\),按关键字排好序,双指针扫一遍,当满足位置\(1,\ldots,

Android 开发技术周报 Issue#283

新闻 Android 11特性调整:安装外部来源应用需要重启APP Google Messages beta版迎来一个新的搜索框 Android开发者生态永远比不上iOS?“联盟与公约”们正改变这一点 谷歌新款Android TV串流设备外形曝光:代号Sabrina Android 11 Beta引入诸多改进:3种图标形状 Pixel Launche

Es的一个gui, 感觉还不错

https://github.com/appbaseio/dejavu#easily-connect-and-remember-indices license也比较友好