Copy code

题目：
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

题目如上，这道题目比较简单哈，拿来给大家练练手
代码格式：
请将自己的代码写入到标签 [ code] [ /code]中（注：code前面是没有空格的，因为网页显示的问题，我加了一个空格），将自己的代码加密至少100以上，违者将会扣除FY的哈，所以请注意

最后的裁定还是按大家程序的运行时间和空间来确定，时间越短，空间越小越好
参与者每人10FY
第一名50
第二名40
第三名30

PS:搞ACM的就不要参加了，参加了我也不加FY，我晓得那些人的哈

对不起!您没有登陆,请先登录论坛.

对不起!您没有登陆,请先登录论坛.

Copy code

// Solution by Richard Leon in old programming style
// Ad-hoc problem
#include <stdio.h>
#include <string.h>
#include <ctype.h>

char num[100000][8];
char phone[16];
char trans[26]={'2','2','2','3','3','3','4','4','4','5','5','5','6','6','6','7','7','7','7','8','8','8','9','9','9','9'};

void insertion(int l,int r)
{
  int i,j;
  char temp[8];

  for (i=l+1;i<=r;++i) {
    j=i;
    memcpy(temp,num[j],8);
    while (j>l && strcmp(temp,num[j-1])) {
      memcpy(num[j],num[j-1],8);
      --j;
    }
    memcpy(num[j],temp,8);
  }
}

void quick_sort(int l,int r)
{
  int i,j;
  char temp[8];

  /*
  if (r-l<10) {
    insertion(l,r);
    return;
  }
  */
  if (r<=l) return;
  i=l-1;
  j=r;

  while (i<j) {
    while (strcmp(num[++i],num[r])<0);
    while (strcmp(num[--j],num[r])>0);
    if (i>=j) break;
    memcpy(temp,num[i],8);
    memcpy(num[i],num[j],8);
    memcpy(num[j],temp,8);
  }
 
  memcpy(temp,num[i],8);
  memcpy(num[i],num[r],8);
  memcpy(num[r],temp,8);
 
  quick_sort(l,i-1);
  quick_sort(i+1,r);
}

void print_regular(int p)
{
  putchar(num[p][0]);
  putchar(num[p][1]);
  putchar(num[p][2]);
  putchar('-');
  putchar(num[p][3]);
  putchar(num[p][4]);
  putchar(num[p][5]);
  putchar(num[p][6]);
}

int main()
{
  int n;
  int i,j,k;
  int cnt;
  int found=0;

  scanf("%d",&n);

  for (i=0;i<n;++i) {
    scanf("%s",&phone);
    j=0;k=0;
    while (phone[j]) {
      if (phone[j]!='-') {
    if (isdigit(phone[j])>0)
     num[i][k]=phone[j];
    else if (isupper(phone[j]))
     num[i][k]=trans[phone[j]-'A'];
    k++;
      }
      j++;
    }
    num[i][k]='\0';
  }
  quick_sort(0,n-1);
 
  i=0;
  while(i<n) {
    cnt=1;
    for (j=i+1;j<n;++j) {
      if (strcmp(num[i],num[j])!=0) break;
      ++cnt;
    }
    if (cnt>1) {
      print_regular(i);
      printf(" %d\n",cnt);
      found=1;
    }
    i=j;
  }

  if (found==0)
    printf("No duplicates.\n");
  return 0;
}