poj1002

487-3279

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 26   Accepted Submission(s) : 4

Problem 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

124873279ITS-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

 

 这道题目，看起来很像使手机键盘的设置题目，但是在处理的时候很需要住以操作所花费的时间，以及各种操作的优化，题目告诉的意思很容易让人想到用map<string int>来存储这些数值，以及计数，值得注意的是在输出的时候如果没有大于1输出No duplicates.

代码如下

#include <iostream>
#include <string>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
const int maxn=100100;
map<string,int> sl;
char a[26]={'2','2','2','3','3','3','4','4','4','5','5','5','6','6','6','7',' ','7','7','8','8','8','9','9','9',' '};
string turn(char s[])//将所有的串转化为电话号码的形式的串
{
   int i,k=0;
   string l;
   char a1[maxn];
   for(i=0;i<(int)strlen(s);i++)
   {
       if(s[i]=='Q'||s[i]=='Z'||s[i]=='-')
        continue;
       else if(s[i]>='A'&&s[i]<='Z')
            a1[k++]=a[s[i]-'A'];
        else if((s[i]>='0')&&(s[i]<='9'))
                a1[k++]=s[i];
   }
   a1[k]='\0';
   int p=(int)strlen(a1);//在串的第三个位置插入'-'
        for(int j=p;j>2;j--)
        {
            a1[j]=a1[j-1];
            if(j==3)
            {
                a1[j]='-';
            }
        }
    a1[p+1]='\0';
   l=a1;
   return l;
}
int main()
{
    int n,i,flag=0,k1=0;
    string ph[maxn];
    char b[maxn];
   scanf("%d",&n);
   string::iterator it;
   for(i=0;i<n;i++)//输入对所有的字符串
     {
        scanf("%s",b);
        ph[i]=turn(b);
     }
     for(i=0;i<n;i++)//读入map当中
       {
           sl[ph[i]]++;
       }
    map<string,int>::iterator itl;
     for(itl=sl.begin();itl!=sl.end();itl++)
     {
         k1++;
         if((*itl).second>1)
            cout<<(*itl).first<<" "<<(*itl).second<<endl;
            else
                flag++;
     }
     if(flag==k1)
        printf("No duplicates.\n");
    return 0;
}
/*
5
1234567
4567893
1245698
4568931
5689741
*/