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##The Fool 题目链接 ###Problem Description The Fool is numbered 0 – the number of unlimited potential –and therefore does not have a specific place in the sequence of the Tarot cards. The Fool can be placed either at the beginning of the Major Arcana or at the end. The Major Arcana is often considered as the Fool’s journey through life and as such, he is ever present and therefore needs no number. Given n ∈ N+, print the parity of ∑i=1N [ni]， where [x] = max a (a∈Z,a≤x) ###Input The first line of the input contains one integer T ≤ 100, denoting the number of testcases. Then T testcases follow. In

Research Guide for Video Frame Interpolation with Deep Learning This blog is from: https://heartbeat.fritz.ai/research-guide-for-video-frame-interpolation-with-deep-learning-519ab2eb3dda In this research guide, we’ll look at deep learning papers aimed at synthesizing video frames within an existing video. This could be in between video frames, known as interpolation, or after them, known as extrapolation . The better part of this guide will cover interpolation. Interpolation is useful in software editing tools as well as in generating video animations. It can also be used to generate clear

传送门：> Here < 题意：有N个弹力装置，第i个弹力装置能够把绵羊从i弹到i+k[i]，如果i+k[i]仍然在N之内则接着弹，如果超出N则绵羊被弹飞。现有两种询问：1. 输出从i位置弹几次被弹飞 2.把弹力装置i的k[i]修改为y 解题思路 考虑用LCT来维护弹力装置之间的关系。如果第$i$个弹力装置的弹力系数为$k[i]$，则$link(i, i+k[i])$。如果$i+k[i] > N$，则需要增加虚拟节点$N+1$，可以形象的用来表示被弹飞 因此从$i$出发几次被弹飞也就可以转化为节点$i$到$N+1$之间路径的长度。要求路径长度只需要动态维护链的长度即可——即$split(x, N+1)$，并且用splay维护size。 对于修改弹力系数，也就等于修改了LCT的构建方式。此时我们需要断开旧边，重新连接新边。先$cut$再$link$即可 Code 　　编号是从0开始的（调了1小时），pushup的时候不要把size[ch[x][0]]打成了ch[x][0]（调了40多分钟） /* By DennyQi */ #include <cstdio> #include <queue> #include <cstring> #include <algorithm> #define r read() #define Max(a,b) (((a)>(b)) ? (a) : (b))

This blog is to explain how to install Tensorflow object detection API in Anaconda in Windows 10 as well as how to train train a convolution neural network to do object detection on your own data set. Steps: 1. Installation and Configuration Install Anaconda First we need to install Anaconda on Windows 10. For specific introduction see this link https://medium.com/@GalarnykMichael/install-python-on-windows-anaconda-c63c7c3d1444 After we have installed Anaconda in windows, lets start to install tensorflow Install Tensorflow open an Anaconda Prompt and run the following code C:\>conda create -

https://study.163.com/provider/400000000398149/index.htm?share=2&shareId=400000000398149 （ 欢迎关注博主主页，学习python视频资源，还有大量免费python经典文章） Before we tackle how to handle them, let’s quickly define what an outlier is. An outlier is any data point that is distinctly different from the rest of your data points. When you’re looking at a variable that is relatively normally distributed, you can think of outliers as anything that falls 3 or more standard deviations from its mean. While this will suffice as a working definition, keep in mind that there’s no golden rule for defining what an outlier is.

Ultimate Guide to WeChat for Business 2019 By Iaroslav Kudritskiy (source :https://rocketbots.io/blog/ultimate-wechat-official-account-for-business-guide) This is the guide for you if you're interested in the Chinese market. Companies all over are interested in creating a presence in China. One of the best ways of doing this is by using WeChat Official Account for Business. This guide is going to be a little more intricate than the others because WeChat is one of the more mysterious messaging apps. This post is designed to help you with your research, we're going to be answering questions like

蒲慕明:写给实验室博士的 Email 著名的华人生物学家蒲慕明先生曾经有一封非常著名的email在网上广为流传，这封email是蒲先生写给自己实验室所有博士生和博士后的，其中的观点我(施一公)完全赞同。这封email写的语重心长，从中可以看出蒲先生的良苦用心。我把这封email转给了我实验室的所有学生。 蒲慕明简介 蒲慕明，1948年10月生，中国科学院院士，美国科学院外籍院士，台湾“中研院”院士。现任中国科学院神经科学研究所所长，中国科学院脑科学与智能技术卓越创新中心主任。1970年毕业于台湾清华大学物理系，1974年于美国Johns Hopkins大学获生物物理学博士学位，1974-1976年在美国普度大学生命科学系从事博士后研究，1976-1985年在美国加州大学艾文分校生物物理系任助理教授、副教授、教授，1985-1988年任耶鲁大学医学院分子神经生物学系教授，1988-1995年任美国哥伦比亚大学生命科学系教授，1995-2000年任美国加州大学圣地亚哥分校Stephen Kuffler讲座教授，2001-2006年任美国加州大学伯克利分校分子与细胞生物学系讲座教授和神经生物学部主任，2006-2013年任该校Paul Licht生物学杰出讲座教授。1999年起任中国科学院神经科学研究所首任及现任所长、神经可塑性研究组组长、高级研究员、博士生导师

补一下昨天没发完的一篇 文中公式若显示不全可左右滑动~ 比较有意思的论文 [1] ，关注的点也是在序列建模的位置信息编码。先前的方法通过引入额外的位置编码，在 embedding 层将词向量和位置向量通过加性编码融合， 但是该种方式每个位置向量是独立训练得到的，并不能建模序列的 order relationship （例如邻接或优先关系），作者将此称为 the position independece problem 。 针对该问题论文提出了一种新的位置编码方式，将独立的词向量替换成自变量为位置的函数，于是单词表示会随着位置的变化而平滑地移动，可以更好地建模单词的绝对位置和顺序信息。 其中， 表示此表中序号为 的单词在位置 时的单词向量， 表示函数集合， 表示单词到函数的映射，展开即为， 为了达到上述要求，函数应该满足以下两个条件： Property 1. Position-free offset transformation 对于任意位置 pos 和 ，存在变换 Transform Transform 满足， 特别地，论文考虑 Transform 为线性变换 Property 2. Boundedness 函数应该是有界的， 接下去，论文证明了满足上述两个条件的解函数形式为， ❝ 贴一下论文给的证明：（看不看无所谓，能用就行 haha） 假设函数 满足上述两个条件，则对于任意位置

今天发现LintCode页面刷新不出来了，所以就转战LeetCode。还是像以前一样，做题顺序：难度从低到高，每天至少一题。 Given a pattern and a string str , find if str follows the same pattern. Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str . Example 1: Input: pattern = "abba" , str = "dog cat cat dog" Output: true Example 2: Input:pattern = "abba" , str = "dog cat cat fish" Output: false Example 3: Input: pattern = "aaaa" , str = "dog cat cat dog" Output: false Example 4: Input: pattern = "abba" , str = "dog dog dog dog" Output: false Notes: You may assume pattern contains only

public int upper_bound_ (int n, int v, int[] a) { if(a[n-1]<=v ) return n+1; // write code here int begin=0; int end=n-1; int mid=(end+begin)/2; while(begin<end){ if(a[mid]>=v){ end=mid; }else{ begin = mid+1; } mid=(end+begin)/2; } return mid+1; } 来源： oschina 链接： https://my.oschina.net/u/3126880/blog/4945469