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…

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…

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…

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…

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…

传送门 Time Limit: 1000MS   Memory Limit: 131072KB   64bit IO Format: %lld & %llu Description 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 dictionar…

1212: H.数7 时间限制: 1 Sec  内存限制: 64 MB 提交: 8  解决: 5 标签提交统计讨论版 题目描述 数7是一个简单的饭桌游戏,有很多人围成一桌,先从任意一人开始数数,1.2.3……那样数下去,逢到7的倍数(7.14.21……)和含有7的数字(17.27……)必须以敲桌子代替.如果有谁逢7却数出来了,就要接受惩罚.小明觉得这个游戏太简单了,于是对它做出了改进,那就是每逢到素数的时候就以敲桌子代替,并且数数的方向发生改变,而且最开始的那个人可以从1到10000中选一个合数…

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…