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 }