Ultraviolet A 755

Question

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 cocould 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 shoshould 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

1124873279ITS-EASY888-45673-10-10-10888-GLOPTUT-GLOP967-11-11310-GINOF101010888-1200-4-8-7-3-2-7-9-487-3279

Sample output

310-1010 2487-3279 4888-4567 3

Ideas and instructions

Basic questions, which can be solved through the Table query directly. parsenum should also be able to compress the time.

Code

#include <iostream>#include <cstring>#include <string>#include <cstring>#include <fstream>#include <sstream>#include <vector>#include <cstdlib>#include <cstdio>#include <map>using namespace std;#ifdef DEBUGifstream in;ofstream out;#endif#ifdef DEBUG#define inpath      "./in.txt"#define outpath     "./out.txt"#define CIN     in#define COUT    out#else#define CIN     cin#define COUT    cout#endif/*A,B,andC mapto   2D,E,and F mapto   3G,H,and I mapto   4J,K,and L mapto   5M,N,and O mapto  6P,R,and S mapto   7T,U,and V mapto   8W,X,and Y mapto   9*/struct keyTable{    keyTable(){            memset(table, 0, sizeof(char) * 256);            table[‘A‘] = ‘2‘;             table[‘B‘] = ‘2‘;              table[‘C‘] = ‘2‘;               table[‘D‘] = ‘3‘;                table[‘E‘] = ‘3‘;                 table[‘F‘] = ‘3‘;                  table[‘G‘] = ‘4‘;                   table[‘H‘] = ‘4‘;                    table[‘I‘] = ‘4‘;                     table[‘J‘] = ‘5‘;                      table[‘K‘] = ‘5‘;                       table[‘L‘] = ‘5‘;                       table[‘M‘] = ‘6‘;                       table[‘N‘] = ‘6‘;                       table[‘O‘] = ‘6‘;                       table[‘P‘] = ‘7‘;                       table[‘R‘] = ‘7‘;                       table[‘S‘] = ‘7‘;                       table[‘T‘] = ‘8‘;                       table[‘U‘] = ‘8‘;                       table[‘V‘] = ‘8‘;                       table[‘W‘] = ‘9‘;                       table[‘X‘] = ‘9‘;                       table[‘Y‘] = ‘9‘;    }    char convert(const char k){            char ret = table[k];            if( 0 == ret )                    return k;            else                    return ret;    }    int table[256];};string parseNum(string vnum,struct keyTable &kt){                string  num;                int idx = 1;                for(string::iterator iter= vnum.begin(); iter != vnum.end(); iter++){                        if( *iter == (char)‘-‘)                               ;                        else{                                num += kt.convert(*iter);                                if(3 == idx)                                    num += ‘-‘;                                  idx++;                        }                }#ifdef DEBUG            cout << vnum<< " -> " << num << endl;#endif // DEBUG            return num;}int main(){#ifdef DEBUG    in.open(inpath,     ios::in);    out.open(outpath,   ios::out);    if(in.fail()){            cout << "input init fail " << "\n";            return -1;    }    if(out.fail()){             cout << "output init fail " << "\n";            return -1;    }#endif    struct keyTable kt;    int cases;    int numTot;    string str;    map<string, int> ansTable;     CIN >> cases;     while(cases--){        ansTable.clear();        CIN >> numTot;        for(int numCurr = 1; numCurr <= numTot; numCurr++){                CIN >> str;                ansTable[parseNum(str, kt)]++;        }        int  flag = 0;        for(map<string, int>::iterator iter = ansTable.begin(); iter != ansTable.end(); iter++){                    if(iter->second > 1){                        COUT << iter->first << " " << iter->second << endl;                        flag = 1;                    }        }        if(0 == flag){                COUT<<"No duplicates."<<endl;        }        if(0 != cases)                COUT << endl;     }return 0;}