1071 Speech Patterns (25 分) People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's iden…

1071 Speech Patterns (25 分)   People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's id…

1071 Speech Patterns People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, w…

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when v…

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when v…

题意: 输入一行字符串,输出出现过次数最多的由字母和数字组成的字符串以及它出现的次数(对大小写不敏感,输出全部输出小写). AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; string word; map<string,int>mp; string s; int main(){ ios::sync_with_stdio(false); cin.t…

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when v…

简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<iostream> #include<algorithm> using namespace std;…

分割字符串的用法+map映射给出input中出现次数最多的单词,如果次数相同,给出按字典序最小的. 这里我用了自定义分隔符来读取字符串,方法如下: //按照定义的分隔符d来分割字符串,对str进行读取 const char *d = "\n\'\"\? ";//换行,',",?,空格 char *p; p = strtok(str,d); //这样做的主要原因是避免最后输出换行 while(p) { //对p进行处理 p=strtok(NULL,d); //读取下一…

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when v…

又是考输入输出 #include <cstdio> #include <cstdlib> #include <string> #include <vector> #include <unordered_map> #include <algorithm> using namespace std; ]; bool is_alphanumerical(char &ch) { ') return true; if (ch >=…

#include <bits/stdc++.h> using namespace std; bool check(char c)//检查是否为字母或数字 { if(c>='A'&&c<='Z'||c>='a'&&c<='z'||c>='0'&&c<='9') return true; else return false; } int main() { map<string,int> count; s…

1071 Speech Patterns (25 分)   People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's id…

1071. Speech Patterns (25) 时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 HOU, Qiming People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops&qu…

Source: PAT A1071 Speech Patterns (25 分) Description: People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to na…

有小伙伴近期问了小编一个问题,说客户需要25*25大小的QR Code二维码,用BarTender怎么做出来?想要指定条形码的大小,还得BarTender符号与版本选项来帮忙.本文小编就来给大家详细讲讲符号版本. 在BarTender中,符号版本指的是条形码的大小.较大的符号版本可以比较小的符号版本容纳更多的数据. 使用“BarTender符号与版本”选项可以设置您的符号的大小,其中大小是按照从最小到最大的顺序列出.如果选择“自动”,BarTender 会自动选择将容纳数据的最小符号版本大小.…

1 /*题目:正整数n若是其平方数的尾部,则称n为同构数 2 如:5*5=25, 25*25=625 3 问: 求1~99中的所有同构数 4 */ 5 //分析:将1-99分为1-9和10-99,用取余的方法得到位数,再判断是否相等 6 7 public class Question4 { 8 public static void main(String[] args) { 9 System.out.println("1-99范围内的同构数如下:"); 10 //for循环遍历1~99…

一.技术总结 开始拿到这道题目时,思考的是我该如何区分它们每一个单词,不知道这里还是要学习得知在cctype头文件中有一个函数用于查看是否为0~9.a~z.A~Z,就是isalnum(),又因为题目中要求不区分大小写,有一个函数tolower(),toupper()要学会合理利用. 然后就是使用map技术了,默认初始化为0如果是string,int,可以直接mp[]++; 然后就是键值和值mp->first,mp->second; 二.参考代码 #include<iostream>…

题目要求的Word定义 Here a "word" is defined as a continuous sequence of alphanumerical characters separated by non-alphanumerical characters or the line beginning/end.  所以 can""can""can 中应该是3个单词 #include <iostream> #include &l…

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when v…

/* 1010. 一元多项式求导 (25) 设计函数求一元多项式的导数.(注:x^n(n为整数)的一阶导数为n*x^n-1.) 输入格式: 以指数递降方式输入多项式非零项系数和指数(绝对值均为不超过1000的整数). 数字间以空格分隔. 输出格式: 以与输入相同的格式输出导数多项式非零项的系数和指数.数字间以空格分隔,但结尾不能有多余空格. 注意“零多项式”的指数和系数都是0,但是表示为“0 0”. 输入样例: 3 4 -5 2 6 1 -2 0 输出样例: 12 3 -10 1 6 0 */…

/* 1005. 继续(3n+1)猜想 (25) 卡拉兹(Callatz)猜想已经在1001中给出了描述.在这个题目里,情况稍微有些复杂. 当我们验证卡拉兹猜想的时候,为了避免重复计算,可以记录下递推过程中遇到的每一个数.例如对n=3进行验证的时候,我们需要计算3.5.8.4.2.1,则当我们对n=5.8.4.2进行验证的时候,就可以直接判定卡拉兹猜想的真伪,而不需要重复计算,因为这4个数已经在验证3的时候遇到过了,我们称5.8.4.2是被3“覆盖”的数.我们称一个数列中的某个数n为“关键数”,…

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when v…

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when v…

题意 输出给定字符串出现最多的字符串(小写输出)和出现次数. 所求字符串要求:字符中可以含有A-Z.0-9. 比如说题目给出的Can1,我们可以转换成can1,can1就算一个字符串整体,而不是单独的取出can1里面的can来加一. 思路 先把大写字母全部转换成小写,然后再用map存储单词个数即可,简单用法. 注意 有个特殊的数据要判断,不写的话牛客可以AC,但是PTA最后一组数据会报错. 特例:给出的字符串最后一个字符如果不是'0'-'9'.'a'-'z'的话,单独算一个字符串判断. for循…

pat 1069 The Black Hole of Numbers 水题,代码如下: #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; bool isSame(char buf[]) { ; while(buf[i] != '\0') { ])return false; else i++; } return true;…

本文为PAT甲级分类汇编系列文章. 线性类,指线性时间复杂度可以完成的题.在1051到1100中,有7道: 题号 标题 分数 大意 时间 1054 The Dominant Color 20 寻找出现最多的数 200ms 1061 Dating 20 寻找字符串中相同字符 200ms 1071 Speech Patterns 25 寻找出现最多的单词 300ms 1077 Kuchiguse 20 字符串共同后缀 150ms 1082 Read Number in Chinese 25 中文读数…

今天开个坑,分类整理PAT甲级题目(https://pintia.cn/problem-sets/994805342720868352/problems/type/7)中1051~1100部分.语言是modern C++. 为什么要整理呢,因为我2019年9月要考PAT甲级,虽然是第一次考,虽然只学了数据结构(https://mooc.study.163.com/course/1000033001?tid=2402970002#/info),但我要冲着高分(2019年9月8日更新:满分)去. 下…

1051 Pop Sequence (25分) [stack] 简答的栈模拟题,只要把过程想清楚就能做出来. 扫描到某个元素时候,假如比栈顶元素还大,说明包括其本身的在内的数字都应该入栈.将栈顶元素和序列比对即可,相同则弹栈,继续扫描:否则无法生成满足条件的序列.注意栈满时不能入栈 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn =…

考虑到PAT甲级考试和开学后的XCPC比赛,决定寒假把PAT (Advanced Level) Practice刷完,进度条会在这篇博客下更新.由于主要以记录为主,大体上不会像单篇题解那么详细,但是对问题的思考,代码的简洁性.可读性还是有保障的,欢迎看到的小伙伴和我讨论 2021.1.10 1001 A+B Format (20分) 很久没写题了,没想到卡了半个小时,惭愧.这里是要把结果用逗号分隔成三组,即以千为单位,不足的话则不必要填逗号,我最多只添了一个逗号,要看清题目意思再动笔 #incl…