Programming Marathon: Phone number __ Programming

Topic Description
The picture above is a telephone of nine Sudoku, as you can see a number corresponding to some letters, so in foreign companies like to design the phone number to match their company name. For example, the company's help desk number is 4357 because 4 corresponds to H, 3 corresponds to E, 5 corresponds to L, 7 corresponds to P, so 4357 is help. Similarly, Tut-glop on behalf of 888-4567, 310-gino representative 310-4466.
Nowcoder just entered the foreign company, not accustomed to this way of naming, now give you a list of phone numbers, please help him convert to digital form of the number, and remove the duplicate parts.

Enter a description:

Input contains multiple sets of data. The

first row of each set of data contains a positive integer n (1≤n≤1024).

followed by n rows, each line contains a phone number, and the phone number consists of a hyphen "-", a number, and a capital letter.
There are no consecutive hyphens, and after the exclusion of hyphens, the length is always 7 (US phone number is only 7 digits).

Output Description:

For each set of inputs, output the standard digital form telephone number that is not duplicated in dictionary order, that is, "xxx-xxxx" form.

one row for each phone number, and a blank line after each set of data as a spacer.

Enter an example:

4873279
its-easy
888-4567
3-10-10-10
888-glop
tut-glop
310-gino
F101010
888-1200
-4-8-7-3-2-7-9-
487-3279
4
utt-help
tut-glop
310-gino
000-1213

Output Example:

310-1010
310-4466
487-3279
888-1200
888-4567
967-1111 000-1213 310-4466 888-4357
888-4567

Write your code here CPP #include <iostream> #include <string.h> #include <map> using namespace std;
	BOOL IsOk (char *str) {int count = 0;

	int n = strlen (str);
	char *p = str;
		while (*p!= ' ") {if (*p = = ') count++;
	p++;
	if (N-count!= 7) return false;
return true;
	BOOL Isnum (char ch) {return (CH-' 0 ') >= 0 && (CH-' 0 ') <= 9;} int main () {int n; 
	Char inputstr[1024];
	
	Char strbuff[256] = {0};
	strbuff[' A '] = ' 2 ';
	strbuff[' B '] = ' 2 ';
	strbuff[' C '] = ' 2 ';
	strbuff[' D '] = ' 3 ';
	strbuff[' E '] = ' 3 ';
	strbuff[' F '] = ' 3 ';
	strbuff[' G '] = ' 4 ';
	strbuff[' H '] = ' 4 ';
	strbuff[' I '] = ' 4 ';
	strbuff[' J '] = ' 5 ';
	strbuff[' K '] = ' 5 ';
	strbuff[' L '] = ' 5 ';
	strbuff[' M '] = ' 6 ';
	strbuff[' N '] = ' 6 ';
	strbuff[' O '] = ' 6 ';
	strbuff[' P '] = ' 7 ';
	strbuff[' Q '] = ' 7 ';
	strbuff[' R '] = ' 7 ';
	strbuff[' S '] = ' 7 ';
	strbuff[' T '] = ' 8 ';
	strbuff[' U '] = ' 8 ';
    strbuff[' V '] = ' 8 ';
	strbuff[' W '] = ' 9 '; 
	strbuff[' X '] = ' 9 ';
	strbuff[' Y '] = ' 9 '; Strbuff[' Z '] = ' 9 ';
	strbuff['-'] = '-';
	CIN >> N;
		while (cin>>n) {map<string, int> MP;
			while (n--) {cin >> inputstr;
			string S;
				if (IsOk (INPUTSTR)) {char *p = INPUTSTR;
						while (*p!= ' ") {if (s.size () = = 3) {s + = '-';
					Continue
						else {if (Isnum (*p)) {s + = *p;
						else {if (*p!= '-') s + = strbuff[*p];
				}} p++;
			} mp.insert (Pair<string, int> (S, 0));
	} map<string, int>:: Iterator it = Mp.begin ();
		while (it!= mp.end ()) {cout << (*it). FIRST.C_STR () << Endl;
	it++;
	} cout << Endl;
return 0; }