短对话 M: Why do you declare the news that you're pregnant on your blog directly? W: I'm so excited that I want to share this good news with the people who love me, but I also hope all my fans can respect my privacy at this special time. Q: How did the…

论文选读二:Multi-Passage Machine Reading Comprehension with Cross-Passage Answer Verification 目前,阅读理解通常会给出一段背景资料,据此提出问题,而问题的答案也往往在背景资料里.不过背景资料一般是一篇文章,或者是文章的一个段落.而对于多篇文章,特别是多篇相近文章时,当前的模型效果就不那么明显了.本文即针对此问题提出的解决方案.此文提出的模型包含三个部分:答案提取模块,答案评价模块,与答案交叉验证模块. 本文提出一个…

Attention-over-Attention Neural Networks for Reading Comprehension 论文地址:https://arxiv.org/pdf/1607.04423.pdf 0 摘要 任务:完形填空是阅读理解是挖掘文档和问题关系的一个代表性问题. 模型:提出一个简单但是新颖的模型A-O-A模型,在文档级的注意力机制上增加一层注意力来确定最后答案 (什么是文档级注意力?就是每阅读问题中的一个词,该词对文档中的所有单词都会形成一个分布,从而形成文档级别的分…

题目链接:https://vjudge.net/problem/HDU-4990 Reading comprehension Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2329    Accepted Submission(s): 954 Problem Description Read the program below care…

题目链接: Reading comprehension Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others) Problem Description   Read the program below carefully then answer the question.#pragma comment(linker, "/STACK:1024000000,1024000000&quo…

论文地址为:Cognitive Graph for Multi-Hop Reading Comprehension at Scale github地址:CogQA 背景 假设你手边有一个维基百科的搜索引擎,可以用来获取实体对应的文本段落,那么如何来回答下面这个复杂的问题呢? “谁是某部在2003年取景于洛杉矶Quality cafe的电影的导演?” 很自然地,我们将会从例如Quality cafe这样的“相关实体”入手,通过维基百科查询相关介绍,并在其中讲到好莱坞电影的时候迅速定位到“Old S…

标题:Neural Machine Reading Comprehension: Methods and Trends 作者:Shanshan Liu, Xin Zhang, Sheng Zhang, Hui Wang, Weiming Zhang 链接:https://arxiv.org/pdf/1907.01118.pdf 摘要:过去几年里,随着深度学习的出现,机器阅读理解(其要求机器基于给定的上下文回答问题)已经赢得了越来越广泛的关注.虽然基于深度学习的机器阅读理解研究正蓬勃发展,但却没有…

Description Read the program below carefully then answer the question. #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include<iostream> #include <cstring> #include <cmath> #include <algorith…

快速幂 #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; long long n,MOD; long long cal(long long a,long long b,long long mod) { ; ) { ==) c=(c*a)%mod,b--; ; } return…

Read the program below carefully then answer the question. #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include<iostream> #include <cstring> #include <cmath> #include <algorithm> #inclu…