487-3279
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 306663 | Accepted: 54796 |
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.
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.
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
题目大致意思：
根据字母到数字的映射关系，找出重复的号码，并且按照字典序输出，并输出重复次数，若没有重复，输出No duplicates.
解法一：map G++过不去 C++能过
主要是利用map的迭代器可以省时间
map 可以自定义比较，插入时候就按照自定义顺序比较的
it->first :key
it->second:value
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<set>
#include<cstdlib>
#include<stdio.h>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
int n;
typedef pair<string,int> PAIR;
//ostream& operator<<(ostream& out,const PAIR &p){
// if(p.second > 1){
// return out << p.first <<" " << p.second;
// }
// //return out<<p.first << "\t"<<p.second;
//}
map<string,int,less<string> > res;
vector<string> pre;
set<string> vis;
bool islet(char c){
if(c >= 65 && c <= 90 || c >= 97 && c <= 122){
return true;
}
return false;
}
void change(string &str){
int len = str.size();
for(int i = 0; i < len; ++i){
if(isdigit(str[i]) || str[i] == '-'){
continue;
}
else{
if(str[i] == 'A' || str[i] == 'B' || str[i] == 'C')
str[i] = '2';
else if(str[i] == 'D' || str[i] == 'E' || str[i] == 'F')
str[i] = '3';
else if(str[i] == 'G' || str[i] == 'H' || str[i] == 'I')
str[i] = '4';
else if(str[i] == 'J' || str[i] == 'K' || str[i] == 'L')
str[i] = '5';
else if(str[i] == 'M' || str[i] == 'N' || str[i] == 'O')
str[i] = '6';
else if(str[i] == 'P' || str[i] == 'R' || str[i] == 'S')
str[i] = '7';
else if(str[i] == 'T' || str[i] == 'U' || str[i] == 'V')
str[i] = '8';
else if(str[i] == 'W' || str[i] == 'X' || str[i] == 'Y')
str[i] = '9';
}
}
++res[str];
}
int main()
{
ios::sync_with_stdio(false);
string str,t;
cin >> n;
while(n--){
cin >> str;
int len = str.size();
t = "";
int cnt = 0;
for(int i = 0; i < len; ++i){
if(islet(str[i]) || isdigit(str[i])){
t += str[i];
++cnt;
if(cnt == 3){
t += "-";
}
}
}
if(t != ""){
pre.push_back(t);
}
}
for(int i = 0; i < pre.size(); ++i){
// cout << "预处理后的数据 " << pre[i] << "len = " << pre[i].size()<<endl;
change(pre[i]);
// cout << "映射后的数据 " << pre[i] << endl;
}
bool flag = false;
for(map<string,int>::iterator it = res.begin(); it != res.end(); ++it){
if(it->second > 1){
flag = true;
cout << it->first << " "<<it->second <<endl;
}
}
// sort(pre.begin(),pre.end());
// for(int i = 0; i < pre.size(); ++i){
// if(res[pre[i]] > 1 && vis.count(pre[i]) == 0){
// flag = true;
// vis.insert(pre[i]);
// cout << pre[i] << " "<<res[pre[i]] <<endl;
// }
// }
if(!flag){
cout << "No duplicates.\n";
}
return 0;
}
方法二：转化为数字串处理：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<set>
#include<cstdlib>
#include<stdio.h>
#include<cmath>
#include<string>
#include<algorithm>
using namespace std;
int b[100000+5];
int main()
{
int n;
scanf("%d",&n);
getchar();
int cnt = 0;
while(n--){
char a[105];
gets(a);
int len = strlen(a);
int num = 0;
for(int i = 0; i < len; ++i){
if(a[i] == '-')
continue;
if(a[i] >= '0' && a[i] <='9'){
num = num*10 + a[i]-'0';
}
else if(a[i] <= 'O'){
num = num*10 + (a[i]-'A')/3+2;
}
else if(a[i] == 'P' || a[i] == 'R' || a[i] == 'S'){
num = num*10 + 7;
}
else if(a[i] == 'T' || a[i] == 'U' || a[i] == 'V'){
num = num*10 + 8;
}
else if(a[i] == 'W' | a[i] == 'X' || a[i] == 'Y'){
num = num*10 + 9;
}
}
b[cnt++] = num;
}
sort(b,b+cnt);
bool flag = true;
for(int i = 1; i < cnt;){
int num = 1;
while(b[i-1] == b[i] && i < cnt){
++num;
++i;
}
if(num > 1){
flag = false;
printf("%03d-%04d %d\n",b[i-1]/10000,b[i-1]%10000,num);
}
++i;
}
if(flag){
printf("No duplicates.\n");
}
return 0;
}