暑假集训第三周 STL G - 487-3279

G - 487-3279

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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

分析：

这是典型的map映射问题，经过一天的折磨，感觉已经有些入门了，刚做的时候花费了我好长的时间。

刚开始时就因为把w写成了s，造成了总是闪退的后果，这次更是给我了一个深刻的教训，不认真会让你效率变得非常低，以后做题都要静下心来，好好去思考，然后敲代码。

还有一个比较新的知识：迭代器，类似于指针一样，就是把map容器从头遍历到尾，找到所有符合条件的数据。适合多组数据的处理。

字符串（string ）c用c.length来计算长度

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#include<stdio.h>
#include<algorithm>
#include<string>
#include<iostream>
#include<string.h>
#include<map>
using namespace std;
map<string,int> g;
int main()
{
    int i,j,k,n;
    char b[100];
    string str,c;
    scanf("%d",&n);
    for(k=0; k<n; k++)
    {
        j=0;
        memset(b,0,sizeof(b));
        cin>>c;
        for(i=0; i<c.length(); i++)
        {
            if(c[i]=='-')
                continue;
            if(c[i]=='A'||c[i]=='B'||c[i]=='C')
                b[j++]='2';
            if(c[i]=='D'||c[i]=='E'||c[i]=='F')
                b[j++]='3';
            if(c[i]=='G'||c[i]=='H'||c[i]=='I')
                b[j++]='4';
            if(c[i]=='J'||c[i]=='K'||c[i]=='L')
                b[j++]='5';
            if(c[i]=='M'||c[i]=='N'||c[i]=='O')
                b[j++]='6';
            if(c[i]=='P'||c[i]=='R'||c[i]=='S')
                b[j++]='7';
            if(c[i]=='T'||c[i]=='U'||c[i]=='V')
                b[j++]='8';
            if(c[i]=='W'||c[i]=='X'||c[i]=='Y')
                b[j++]='9';
            if(c[i]>='0'&&c[i]<='9')
                b[j++]=c[i];
                if(j==3)
                    b[j++]='-';
        }
        str=b;
        g[str]++;
    }
    map<string,int>::iterator p;  /遍历查找
    int flag=0;
    for(p=g.begin(); p!=g.end(); p++)
    {
        if(p->second>1)
        {
            flag=1;
            str=p->first;
           cout<<str<<" "<<p->second<<endl;
        }
    }
    if(flag==0)
        printf("No duplicates.\n");
    return 0;
}