C -- POJ 1002 487-3279

487-3279

Time Limit: 2000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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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

Source

East Central North America 1999
 
题目大意：
 
把一个字符串转化成固定格式的电话号码，输出有重复的号码及个数，即若该号码只出现一次不输出，出现一次以上输出。
 
题目分析：
 
用STL中的map。首先把字符串转化成电话号码，用map把号码和出现的次数进行映射。需要注意的是，这个题目数据量变态大，特别容易超时，看题解都说直接处理字符串会超时，所以都把转化都的号码变成整形，但是，我依然超时了，最后发现是输入的字符串数组开的太小（20），后来改成（50）就过了，我也是不太懂为什么。
 
代码：

1 #include <iostream>
2 #include <cstdio>
3 #include <map>
4 #include <string.h>
5 using namespace std;
6 int fun(char c){
7     switch(c){
8     case 'A':
9     case 'B':
10     case 'C': return 2;
11     case 'D':
12     case 'E':
13     case 'F': return 3;
14     case 'G':
15     case 'H':
16     case 'I': return 4;
17     case 'J':
18     case 'K':
19     case 'L': return 5;
20     case 'M':
21     case 'N':
22     case 'O': return 6;
23     case 'P':
24     case 'R':
25     case 'S': return 7;
26     case 'T':
27     case 'U':
28     case 'V': return 8;
29     case 'W':
30     case 'X':
31     case 'Y': return 9;
32     }
33 }
34 int main(){
35     map<int ,int> mp;
36     mp.clear();
37     int n;
38     scanf("%d",&n);
39     while(n--){
40        int num=0;
41        char a[50];
42        scanf("%s",a);
43        for(int i=0;i<strlen(a);i++){
44         if(a[i]=='-') continue;
45         else if(a[i]>='A'&&a[i]<='Z') num=num*10+fun(a[i]);
46         else num=num*10+a[i]-'0';
47        }
48        mp[num]++;
49     }
50     bool flag=true;
51     for(map<int ,int >::iterator it=mp.begin();it!=mp.end();it++){
52         if(it->second>1){
53             printf("%03d-%04d %d\n",it->first/10000,it->first%10000,it->second);
54             flag=false;
55         }
56     }
57     if(flag) cout<<"No duplicates."<<endl;
58     return 0;
59 }