• uva 755


     1 #include <iostream>
     2 #include <string>
     3 #include <map>
     4 #include <algorithm>
     5 #include <cstdio>
     6 #include <cctype>
     7 using namespace std;
     8 
     9 const char kTable[] = "2223334445556667Q77888999Z";
    10 int main()
    11 {
    12     int T;
    13     cin >> T;
    14     bool first_output(true);
    15     while(T--)
    16     {
    17         int n; cin >> n;
    18         map<string, int> r;
    19         for(int i = 1; i <= n; i++)
    20         {
    21             string t; cin >> t;
    22             t.erase(remove(t.begin(), t.end(), '-'), t.end());
    23             for(int j = 0; j < t.size(); j++)
    24                 if(!isdigit(t[j]))
    25                     t[j] = kTable[t[j]- 'A'];
    26             t.insert(3, "-");
    27             r[t]++;
    28         }
    29         if(first_output)
    30             first_output = false;
    31         else cout << endl;
    32         bool found(false);
    33         for(map<string, int>::iterator it = r.begin(); it != r.end(); it++)
    34             if( it->second > 1)
    35             {
    36                 cout << it->first << " " << it->second << endl;
    37                 found = true;
    38             }
    39         if(!found) cout << "No duplicates." << endl;
    40     }
    41     return 0;
    42 }

    另一种解法:

    先將每一種不同格式的電話號碼全部換成7位數整數,
    利用一個hash紀錄每一種電話號碼的出現的次數,
    將出現兩次以上的電話號碼紀錄到一個陣列裡面,
    再利用quicksort將這個陣列以電話號碼來排序,
    最後從頭將電話號碼及其出現的次數輸出來即可。

      1 #include<stdio.h>
      2 #include<string.h>
      3 #include<ctype.h>
      4 
      5 void quicksort( int start, int end, int array[] )
      6 {
      7     if( start < end )
      8     {
      9         int change = start;
     10         int i;
     11         int temp;
     12         for( i = start+1 ; i < end ; i++ )
     13             if( array[i] < array[start] )
     14             {
     15                 change++;
     16 
     17                 temp = array[i];
     18                 array[i] = array[change];
     19                 array[change] = temp;
     20             }
     21 
     22         temp = array[start];
     23         array[start] = array[change];
     24         array[change] = temp;
     25 
     26         quicksort( start, change, array );
     27         quicksort( change+1, end, array );
     28     }
     29 }
     30 
     31 int hash_numbers[10000005] = {0};
     32 
     33 int main()
     34 {
     35     int datasets;
     36     int blank;
     37     while( scanf( "%d", &datasets ) != EOF )
     38     {
     39         blank = 0 ;
     40         int numbers;
     41 
     42         while( datasets-- )
     43         {
     44             scanf( "%d", &numbers );
     45             getchar();
     46 
     47             int i;
     48             memset( hash_numbers, 0, sizeof(hash_numbers) );
     49             int output[100005] = {0};
     50             int output_saved = 0;
     51             for( i = 0 ; i < numbers ; i++ )
     52             {
     53                 char tempnum[1005];
     54                 gets( tempnum );
     55 
     56                 int templen = strlen( tempnum );
     57                 int tempnumint = 0;
     58                 int j;
     59                 for( j = 0 ; j < templen ; j++ )
     60                 {
     61                     if( isalnum( tempnum[j] ) )
     62                     {
     63                         tempnumint *= 10;
     64                         if( isdigit( tempnum[j] ) )
     65                             tempnumint += (int)(tempnum[j] - '0');
     66                         else
     67                         {
     68                             switch( tempnum[j] )
     69                             {
     70                                 case 'A': case 'B': case 'C':
     71                                     tempnumint += 2;
     72                                     break;
     73                                 case 'D': case 'E': case 'F':
     74                                     tempnumint += 3;
     75                                     break;
     76                                 case 'G': case 'H': case 'I':
     77                                     tempnumint += 4;
     78                                     break;
     79                                 case 'J': case 'K': case 'L':
     80                                     tempnumint += 5;
     81                                     break;
     82                                 case 'M': case 'N': case 'O':
     83                                     tempnumint += 6;
     84                                     break;
     85                                 case 'P': case 'R': case 'S':
     86                                     tempnumint += 7;
     87                                     break;
     88                                 case 'T': case 'U': case 'V':
     89                                     tempnumint += 8;
     90                                     break;
     91                                 case 'W': case 'X': case 'Y':
     92                                     tempnumint += 9;
     93                                     break;
     94                             }
     95                         }
     96                     }
     97                 }
     98                 hash_numbers[tempnumint]++;
     99                 if( hash_numbers[tempnumint] == 2 )
    100                     output[output_saved++] = tempnumint;
    101             }
    102             quicksort( 0, output_saved, output );
    103 
    104             if( blank )
    105                 printf( "
    " );
    106             blank = 1;
    107 
    108             for( i = 0 ; i < output_saved ; i++ )
    109                 printf( "%03d-%04d %d
    ", output[i]/10000, output[i]%10000, hash_numbers[output[i]] );
    110             if( output_saved == 0 )
    111                 printf( "No duplicates.
    " );
    112         }
    113     }
    114     return 0;
    115 }
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  • 原文地址:https://www.cnblogs.com/aze-003/p/5161058.html
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