UVa 755 487--3279

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 first line of the input contains the number of datasets in the input. A blank line follows. The first line of each dataset 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. There's a blank line between datasets.

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.

Print a blank line between datasets.

Sample Input 

1

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

Miguel A. Revilla

2000-02-09

输入T，后面跟着T个输入块。

每个输入块第一行输入N，后面跟着N个电话号码，电话号码由数字字母和“-”组成。

通过给出的字母映射表把字母转化为数字，并去除多余的“-”，最后输出的电话格式为xxx-xxxx。

统计相同号码出现的次数，输出重复的电话号码及出现次数，按电话号码的数字排序。

若全部没有重复，输出一行表示没有重复。

每个输入块对应的输出块之间有一个空行。

基本的字符串处理，使用了cctype中判断数字字母的方法。

然后使用map存入，键值为电话号码，值为次数，自带排序。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <cctype> //包含isdigit()和isalpha()方法
using namespace std;

map<string,int> m;
char p[10];
char str[100];

//字母所对应的数字的映射
char let2num[26] = {'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', '9'};

int main(){
//freopen("test.in","rt",stdin);
int t,n,i,j;
cin >>t;
while(t--){
cin >>n;
int flag = 0; //记录是否有重复的号码
m.clear();
for(i = 0;i < n;i++){
cin >>str;
int len = strlen(str);
int cnt = 0;
p[3] = '-';
for(j = 0;j < len;j++){
if(cnt == 3){ //电话号码格式固定第4个为“-”
cnt ++;
}
if(isdigit(str[j])){ //若为数字，则不变
p[cnt++] = str[j];
}
else if(isalpha(str[j]) && (str[j] != 'Q'|| str[j] != 'Z')){ //将除Q，Z外的字母转换为数字
p[cnt++] = let2num[str[j] - 'A'];
}
}
string phone(p); //构造字符串
if(m.count(phone)){ //记录判断次数及将其插入map中
m[phone]++;
}
else{
m[phone] = 1;
}
}
for(map<string,int>::iterator it = m.begin();it != m.end();++it){ //迭代器遍历map输出
if(it->second > 1){ //只输出重复的电话号码，若有重复则flag = 1；
cout <<it->first <<" " <<it->second <<endl;
flag = 1;
}
}
if(!flag){ //当没有重复时输出 No duplicates.cout <<"No duplicates." <<endl;
}
if(t != 0){
4000
//输出块之间的空行
cout <<endl;
}
}
return 0;
}