while director 轉化 mem mov algorithm eas turn between

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

  1 /*
  2 問題 輸入包含大寫字母、-、數字的字符串，將其中的大寫字母轉換成對應的數字，即標準化後，將出現超過2次的號碼標準的形式
  3 按字典序輸出。
  4 解題思路 先將字符串存儲，將其轉化為標準形式存入map映照容器中，其中插入鍵值和映照數據是註意pair的使用，條件輸出即可。
  5 */
  6 #include<cstdio>
  7 #include<string>
  8 #include<cctype>
  9 #include<cstring>
 10 #include<iostream>
 11 #include<vector>
 12 #include<algorithm>
 13 #include<map>
 14 using namespace std;
 15 
 16 struct phoneNumber{
 17     string num;
 18     int cou;
 19 }phoList[100100];
 20 
 21 int phocou;
 22 
 23 string conver(char *str);
 24 char aton(char a);
 25 void check(string str);
 26 
 27 map<string,int> m;
 28 
 29 int main()
 30 {
 31     int n,i;
 32     string temp;
 33     char str[51];
 34 
 35     while(scanf("%d",&n) != EOF){
 36         for(i=0;i<n;i++){
 37             scanf("%s",str);
 38             temp=conver(str);
 39             check(temp);
 40         }    
 41         /*for(i=0;i<phocou;i++){
 42             cout<<phoList[i].num<<‘ ‘<<phoList[i].cou<<endl;
 43         }*/
 44         
 45         int flag=0;
 46         map<string,int>::iterator it;
 47         for(it=m.begin();it != m.end(); it++){
 48             if(it->second > 1){
 49                 flag=1;
 50                 cout<<it->first<<‘ ‘<<it->second<<endl;
 51             }    
 52         }
 53         if(!flag)
 54             cout<<"No duplicates.\n";
 55     }
 56     return 0;
 57 } 
 58 
 59 void check(string str)
 60 {
 61     map<string,int>::iterator it;
 62     it=m.find(str);
 63     if(it == m.end())
 64         m.insert( pair<string,int>(str,1) );
 65     else
 66         it->second++;
 67 }
 68 
 69 string conver(char *str)
 70 {
 71     int len=strlen(str),i;
 72     string sum;
 73     for(i=0;i<len;i++){
 74         if(isalpha(str[i])){
 75             sum = sum + aton(str[i]);    
 76         }
 77         else if(isdigit(str[i])){
 78             sum = sum + str[i];
 79         }
 80     }
 81     string::iterator it=sum.begin();
 82     sum.insert(it + 3,‘-‘);
 83     //cout<<sum<<endl;
 84     return sum;
 85 }
 86 
 87 char aton(char a)
 88 {
 89     if(a == ‘A‘ || a == ‘B‘ || a == ‘C‘)
 90         return ‘2‘;
 91     if(a == ‘D‘ || a == ‘E‘ || a == ‘F‘)
 92         return ‘3‘;
 93     if(a == ‘G‘ || a == ‘H‘ || a == ‘I‘)
 94         return ‘4‘;
 95     if(a == ‘J‘ || a == ‘K‘ || a == ‘L‘)
 96         return ‘5‘;
 97     if(a == ‘M‘ || a == ‘N‘ || a == ‘O‘)
 98         return ‘6‘;
 99     if(a == ‘P‘ || a == ‘R‘ || a == ‘S‘)
100         return ‘7‘;
101     if(a == ‘T‘ || a == ‘U‘ || a == ‘V‘)
102         return ‘8‘;
103     if(a == ‘W‘ || a == ‘X‘ || a == ‘Y‘)
104         return ‘9‘;
105 }

POJ 1002 487-3279(map映照容器的使用)