POJ 1002-487-3279 487-3279 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 245807 Accepted: 43582 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 Source East Central North America 1999
这题的思路其实很简单，使用C++的map搞定。 首先将字符串转换成标准的格式，然后将题目中给的对应关系转换成一个简单的hash数组，将字符串中的字符转换成数字（char），最后加入到map中，加入的方法使用map[s]++，如果s在map中，则s出现的次数加1；反之，则将s加入到map中，然后次数加1，即0++=1。这个方法很巧妙，是我从《C++ Primer》上看到的。 最后就是遍历map，如果s出现次数大于1，则将其输出；如果都不大于1，则输出No duplicates. 代码如下： [cpp] #include<iostream> #include<map> #include<string> using namespace std; char hash_h[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′,’0′};//题目给出的map对应关系，其中q和z对应0 //将输入转换成标准格式 string get_standard_form(string s) { string rs=""; int s_size=s.size(); for(int i=0;i<s_size;i++) { if(s[i]!=’-‘)//将s中的连接符都去掉 rs+=s[i]; } s_size=rs.size(); for(int i=0;i<s_size;i++) { if(rs[i]>=’A’&&rs[i]<=’Z’) rs[i]=hash_h[rs[i]-‘A’];//替换 } return rs.substr(0,3)+"-"+rs.substr(3);//添加连接符 } int main() { int n; string s; cin>>n; map<string,int> msi; while(n–) { cin>>s; msi[get_standard_form(s)]++;//使用map存储 } map<string,int>::iterator it=msi.begin();//充分利用map的特性 int count=0; while(it!=msi.end())//map已经自动按字典序排好了 { if(it->second>1) { cout<<it->first<<" "<<it->second<<endl; count++; } it++; } if(count==0)//记得判断是否有重复的！ cout<<"No duplicates. "<<endl; return 0; } [/cpp] 本代码提交AC，用时1594MS，内存5160K。]]>