POJ-1002-487-3279(字符串)
487-3279
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 309685 Accepted: 55292
Description
Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phrase. For example, you can call the University of Waterloo by dialing the memorable TUT-GLOP. Sometimes only part of the number is used to spell a word. When you get back to your hotel tonight you can order a pizza from Gino's by dialing 310-GINO. Another way to make a telephone number memorable is to group the digits in a memorable way. You could order your pizza from Pizza Hut by calling their ``three tens'' number 3-10-10-10.
The standard form of a telephone number is seven decimal digits with a hyphen between the third and fourth digits (e.g. 888-1200). The keypad of a phone supplies the mapping of letters to numbers, as follows:
A, B, and C map to 2
D, E, and F map to 3
G, H, and I map to 4
J, K, and L map to 5
M, N, and O map to 6
P, R, and S map to 7
T, U, and V map to 8
W, X, and Y map to 9
There is no mapping for Q or Z. Hyphens are not dialed, and can be added and removed as necessary. The standard form of TUT-GLOP is 888-4567, the standard form of 310-GINO is 310-4466, and the standard form of 3-10-10-10 is 310-1010.
Two telephone numbers are equivalent if they have the same standard form. (They dial the same number.)
Your company is compiling a directory of telephone numbers from local businesses. As part of the quality control process you want to check that no two (or more) businesses in the directory have the same telephone number.
Input
The input will consist of one case. The first line of the input specifies the number of telephone numbers in the directory (up to 100,000) as a positive integer alone on the line. The remaining lines list the telephone numbers in the directory, with each number alone on a line. Each telephone number consists of a string composed of decimal digits, uppercase letters (excluding Q and Z) and hyphens. Exactly seven of the characters in the string will be digits or letters.
Output
Generate a line of output for each telephone number that appears more than once in any form. The line should give the telephone number in standard form, followed by a space, followed by the number of times the telephone number appears in the directory. Arrange the output lines by telephone number in ascending lexicographical order. If there are no duplicates in the input print the line:
No duplicates.
Sample Input
12
4873279
ITS-EASY
888-4567
3-10-10-10
888-GLOP
TUT-GLOP
967-11-11
310-GINO
F101010
888-1200
-4-8-7-3-2-7-9-
487-3279
Sample Output
310-1010 2
487-3279 4
888-4567 3
分析
直接看代码吧。也可以用排序做,不过map本身就是按照key排序储存的,利用这一特性直接就可以。
int dit[25] = {2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9};
int n;
string str[100005];
map<string,int> M;
void change(string& str)
{
string tmp = str;
int k = 0;
for(int i=0;i<str.length();i++)
{
if(str[i] == '-')
continue;
if(str[i]>='A'&&str[i]<='Z')
tmp[k++] = dit[str[i]-'A']+'0';
else
tmp[k++] = str[i];
}
tmp.erase(tmp.begin()+k,tmp.end());
tmp.insert(3,"-");
str = tmp;
}
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>str[i];
change(str[i]);
M[str[i]]++;
}
int flag = 0;
for(map<string,int>::iterator it = M.begin();it!=M.end();it++)
{
if(it->second==1)continue;
cout<<it->first<<' '<<it->second<<endl;
flag = 1;
}
if(flag == 0)
cout<<"No duplicates."<<endl;
return 0;
}
POJ-1002-487-3279(字符串)的更多相关文章
- 字符串专题:map POJ 1002
第一次用到是在‘校内赛总结’扫地那道题里面,大同小异 map<string,int>str 可以专用做做字符串的匹配之类的处理 string donser; str [donser]++ ...
- [POJ 1002] 487-3279 C++解题报告
487-3279 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 228365 Accepted: 39826 D ...
- POJ 1002 487-3279
A - 487-3279 Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- Poj 1002 487-3279(二叉搜索树)
题目链接:http://poj.org/problem?id=1002 思路分析:先对输入字符进行处理,转换为标准形式:插入标准形式的电话号码到查找树中,若有相同号码计数器增加1,再中序遍历查找树. ...
- POJ 1002 UVA 755 487--3279 电话排序 简单但不容易的水题
题意:给你许多串字符串,从中提取电话号码,输出出现复数次的电话号码及次数. 以下是我艰难的AC历程:(这题估计是我刷的题目题解次数排前的了...) 题目不是很难理解,刚开始想到用map,但stl的ma ...
- POJ 1035 Spell checker 字符串 难度:0
题目 http://poj.org/problem?id=1035 题意 字典匹配,单词表共有1e4个单词,单词长度小于15,需要对最多50个单词进行匹配.在匹配时,如果直接匹配可以找到待匹配串,则直 ...
- POJ 1200 Crazy Search(字符串简单的hash)
题目:http://poj.org/problem?id=1200 最近看了一个关于hash的问题,不是很明白,于是乎就找了些关于这方面的题目,这道题是一道简单的hash 字符串题目,就先从他入手吧. ...
- 开篇,UVA 755 && POJ 1002 487--3279 (Trie + DFS / sort)
博客第一篇写在11月1号,果然die die die die die alone~ 一道不太难的题,白书里被放到排序这一节,半年前用快排A过一次,但是现在做的时候发现可以用字典树加深搜,于是乐呵呵的开 ...
- poj 2041 Unreliable Message 字符串处理
水的问题.直接附着到代码. //poj 2041 //sep9 #include <iostream> using namespace std; char mode[128]; char ...
- POJ 1002 - 487-3279 STL
先把不是标准格式的字符串变成标准格式再输出出现两次以上的标准串和出现的次数不然输出 "No duplicates." #include <iostream> #incl ...
随机推荐
- Android EditText实现小数点后几位的终级方案
有时候,我们用EditText的时候,会要求输入小数点后几位,遇到过几次这样的需求,这次把它给记下来,方便以后使用 /** * 小数位数 */ public class PointLengthFilt ...
- Educational Codeforces Round 21E selling souvenirs (dp)
传送门 题意 给出n个体积为wi,价值为ci的物品,现在有一个m大的背包 问如何装使得最后背包内的物品价值最大,输出价值 分析 一般的思路是01背包,但n*v不可做 题解的思路 We can iter ...
- Codeforces 711B 【模拟】
比赛的时候绝壁打麻烦了... 考虑的好麻烦...wa7...还要判断出来的是不是positive的... 好吧..认了.. #include<cstdio> #include <ma ...
- 安装 statconn 使R与c#交互
很久以前完成过,但是最近重新折腾发现全忘了,所以记下来. 1.安装 R 2.安装 R studio 3.去 http://rcom.univie.ac.at/download.html 下载 stat ...
- bzoj 3365: [Usaco2004 Feb]Distance Statistics 路程统计【容斥原理+点分治】
统计在一个root下的两个子树,每个子树都和前面的运算一下再加进去对于这种需要排序的运算很麻烦,所以考虑先不去同子树内点对的算出合法点对个数,然后减去每一棵子树内的合法点对(它们实际上是不合法的,相当 ...
- Mybatis 未设置主键映射报错;Cause: java.sql.SQLSyntaxErrorException: Unknown column 'system_id' in 'field list'
使用MyBatis的时候,主键的字段建议绑定在Bean的属性上面, import javax.persistence.*; public class User { @Id @Column(name = ...
- 强连通分量再探 By cellur925
我真的好喜欢图论啊. (虽然可能理解的并不深hhh) 上一次(暑假)我们初探了强联通分量,这一次我们再探.(特别感谢pku-lyc老师的课件.有很多引用) 上次我们忘记讨论复杂度了.tarjan老爷爷 ...
- eclipse中alt+/的作用
一般情况下alt+/有代码提示作用,还有代码提示的快捷代码也不是alt+/,因此要恢复代码提示用alt+/.需要做两件事.在 Window - Preferences - General - Keys ...
- CSS常见的五大布局
本文概要 本文将介绍如下几种常见的布局: 一.单列布局 常见的单列布局有两种: header,content 和 footer 等宽的单列布局 header 与 footer 等宽,content 略 ...
- iOS [CIContext initWithOptions:]: unrecognized selector sent to instance 模拟器 iOS 8.4
在模拟器(iPhone 4s,iOS 8.4)中运行应用时, 应用crash在了使用CIContext(options:nil) 这个API的一个纯Swift第三方库. StackOverFlow的解 ...