zoukankan      html  css  js  c++  java
  • 字母对应数字

    1. /*Description
    2. 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.
    3. 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:
    4. A, B, and C map to 2
    5. D, E, and F map to 3
    6. G, H, and I map to 4
    7. J, K, and L map to 5
    8. M, N, and O map to 6
    9. P, R, and S map to 7
    10. T, U, and V map to 8
    11. W, X, and Y map to 9
    12. 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.
    13. Two telephone numbers are equivalent if they have the same standard form. (They dial the same number.)
    14. 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.
    15. Input
    16. 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.
    17. Output
    18. 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:
    19. No duplicates.
    20. */
    21. #include <iostream>
    22. #include <math.h>
    23. #include <string>
    24. #include <map>
    25. using namespace std;
    26. #include <fstream>
    27. //ifstream fin("C:\Users\yw\Desktop\E\E.2.dat");
    28. //ofstream fout("result1.txt");
    29. //#define cout fout
    30. //#define cin fin
    31. int main()
    32. {
    33. int n;
    34. cin>>n;
    35. //getchar();
    36. map<string,int>result; //存结果对应的map的
    37. map< string,int>::iterator iter; //建立反向迭代器
    38. map<char,int>num;
    39. // char num[91];
    40. num['A' ]=num['B']=num['C']='2'; //字母对应的数字
    41. num['D' ]=num['E']=num['F']='3';
    42. num['G' ]=num['H']=num['I']='4';
    43. num['J' ]=num['K']=num['L']='5';
    44. num['M' ]=num['N']=num['O']='6';
    45. num['P' ]=num['R']=num['S']='7';
    46. num['T' ]=num['U']=num['V']='8';
    47. num['W' ]=num['X']=num['Y']='9';
    48. while(n--)
    49. {
    50. char word[100]; //注意字符可能很多
    51. cin>>word;
    52. char *p=word;
    53. string temp="";
    54. int count=0;
    55. while( (*p)!='') //转换数字
    56. {
    57. if( (*p>='A'&&*p<'Q')||(*p>'Q'&&*p<'Z') )
    58. {
    59. temp+=num[*p];
    60. count++; p++;
    61. if(count==3)
    62. temp+='-';
    63. }
    64. else if( *p>='0'&&*p<='9' )
    65. {
    66. temp+=(*p);
    67. count++; p++;
    68. if(count==3)
    69. temp+='-';
    70. }
    71. else p++;
    72. }
    73. iter=result.find(temp); //查找时候在map里面有
    74. if(iter==result.end() )
    75. result[temp]=1;
    76. else
    77. result[temp]=iter->second+1; //有就数字加一
    78. }
    79. int ok=1;
    80. for(iter = result.begin(); iter != result.end(); iter++) //利用方向迭代器输出数据
    81. {
    82. if(iter->second>1)
    83. {cout<<iter->first<<' '<<iter->second<<endl; ok=0;}
    84. }
    85. if(ok) cout<<"No duplicates."<<endl;
    86. return 0;
    87. }





    附件列表

    • 相关阅读:
      Java面向对象_常用类库api——二分查找算法
      Spyder中代码提示功能添加
      随机数种子random.seed()理解
      Windows10下Anaconda+Tensorflow+Keras环境配置
      面向对象分析与设计—OOD部分
      面向对象分析与设计—OOA部分
      面向对象分析与设计—基本概念部分
      基于聚类K-Means方法实现图像分割
      机器学习中强化学习与监督学习、无监督学习区别
      LeetCode501.二叉搜索树中的众数
    • 原文地址:https://www.cnblogs.com/sober-reflection/p/b4beb4c7ed869062e97c61f55073efc1.html
    Copyright © 2011-2022 走看看