Updating a Dictionary  In this problem, a dictionary is collection of key-value pairs, where keys are lower-case letters, and values are non-negative integers. Given an old dictionary and a new dictionary, find out what were changed. Each dictionar…

全是字符串相关处理,截取长度等相关操作的练习 AC代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cctype> #include <cstring> #include <string> #include <sstream> #include <vector> #include <set> #include…

题目 题目     分析 第一次用stringstream,真TMD的好用     代码 #include <bits/stdc++.h> using namespace std; int main() { int n; cin>>n; getchar();//回车 while(n--) { string s1,s2; getline(cin,s1); getline(cin,s2); for(int i=0;i<s1.length();i++) if(!isalpha(s1…

大致题意:用{ key:value, key:value, key:value }的形式表示一个字典key表示建,在一个字典内没有重复,value则可能重复 题目输入两个字典,如{a:3,b:4,c:10,f:16}  {a:3,c:5,d:10,ee:4} 于新字典相比,找出旧字典的不同,并且输出用相应格式,增加的用'+',如 +d,ee 减少的用'-',如 -b,f value变化的用*,如 *c 没有不同则输出"No changes"" 如果使用java,则可以用Tre…

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1113 1113: Updating a Dictionary Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 491  Solved: 121[Submit][Status][Web Board] Description In this problem, a dictionary is collection of key-value pairs, w…

UVA12504 - Updating a Dictionary 给出两个字符串,以相同的格式表示原字典和更新后的字典.要求找出新字典和旧字典的不同,以规定的格式输出. 算法操作: (1)处理旧字典,将旧字典中的每对关键字及其价值从字典串中截取出来,压入容器中.用做新旧字典对比检索. (2)处理新字典,将新字典中的每对关键字及其价值从字典串中截取出来.对于每个关键字,在容器中检索相同的关键字.若检索不成功,该关键字是新字典新增的,处理存储到相关的串数组中.若检索成功,提取容器中该关键字的价值与新…

题意:对比新老字典的区别:内容多了.少了还是修改了. 代码:(Accepted,0.000s) //UVa12504 - Updating a Dictionary //#define _XieNaoban_ #include<iostream> #include<sstream> #include<string> #include<vector> #include<map> using namespace std; int T; int mai…

Problem C     Updating a Dictionary In this problem, a dictionary is collection of key-value pairs, where keys are lower-case letters, and values are non-negative integers. Given an old dictionary and a new dictionary, find out what were changed. Eac…

In this problem, a dictionary is collection of key-value pairs, where keys are lower-case letters, and values are non-negative integers. Given an old dictionary and a new dictionary, find out what were changed. Each dictionary is formatting as follow…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 不确定某个map里面是否有某个关键字的时候. 要用find来确定. 如果直接用访问下标的形式去做的话. 会强行给他加一个那个关键字(原来没有). (当然那个关键字的映射为空就是了); [代码] #include <bits/stdc++.h> using namespace std; string s; map<string, string> mmap1, mmap2; vector <string>…