487-3279(输入外挂）

http://acm.tju.edu.cn/toj/showp1161.html1161.   487-3279

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Time Limit: 5.0 Seconds   Memory Limit: 65536K
Total Runs: 2106   Accepted Runs: 571    Multiple test files
Case Time Limit: 2.5 Seconds

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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

Hint

The lexicographical order is:
Sort by first character in increasing order (0, 1, 2, ..., 9). If there is a tie, break ties with second character, and so on.

 

题意：根据所给对应法则将输入的字符串对应成电话号码，输出出现两次及以上次数的点换号码

这里因为读入的数据特别的多，所以读入用scanf会超时，而用getchar() 读入一个字符会快很多，所以这里介绍一个输出外挂的方法，

给出代码

#include<iostream>
#include<cstdio>
#include<sstream>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
#define N 100005
int s[26] = {2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,0,7,7,8,8,8,9,9,9};
int ans[N];
int in()//输入外挂
{
    char ch;
    int res = 0;
    while(ch = getchar())
    {
        if(ch=='\n') break;
        if(ch>='0'&&ch<='9')
        {
            res*=10; res+=(ch-'0');
        }
        else if(ch>='A'&&ch<='Z')
        {
            res*=10 ; res+=(s[ch-'A']);
        }
    }
    return res;
}
int main()
{
    int n;
    scanf("%d",&n);
    getchar();//一定要加这句，因为getchar 会将末尾的换行符读进去，影响下次的外挂，所以写外挂习惯性的要在前一次的scanf()后加上getchar
    char dh[80];
    for(int i = 0 ; i < n ; i++)
    {
        ans[i] = in();
    }
    sort(ans,ans+n);
    int flag = false;
    for(int i = 0;i < n ;)
    {
        int c = 1;
        int j = i;
        while(j+1<n&&ans[j]==ans[j+1])
        {
            c++;
            //printf("haha\n");
            j++;
            flag = true;
        }
        char tm[] = "000-0000";
        int r = ans[i];
        for(int j = 7 ; ~j ;j--)//j是-1 的时候~表示按位取反，-1按位取反的值是0
        {
            tm[j] ='0'+ r%10;
            r /= 10;
            if(j==4) j--;
        }
        if(c>1)
        {
              printf("%s %d\n",tm,c);
              //printf(" i = %d\n",i);
        }
        i = i + c ;
    }
    if(!flag) puts("No duplicates.");
    return 0;
}