UVa 755 / POJ 1002 / East Central North America 1999 487--3279 (排序)
755 - 487--3279
Time limit: 3.000 seconds
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=98&page=show_problem&problem=696
http://poj.org/problem?id=1002
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 first line of the input contains the number of datasets in the input. A blank line follows. The first line of each dataset 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. There's a blank line between datasets.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.
Print a blank line between datasets.
Sample Input
1 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
电话号码的标准格式是七位十进制数，并在第三、第四位数字之间有一个连接符。电话拨号盘提供了从字母到数字的映射，映射关系如下：
A, B, 和C 映射到 2
D, E, 和F 映射到 3
G, H, 和I 映射到 4
J, K, 和L 映射到 5
M, N, 和O 映射到 6
P, R, 和S 映射到 7
T, U, 和V 映射到 8
W, X, 和Y 映射到 9
Q和Z没有映射到任何数字，连字符不需要拨号，可以任意添加和删除。 TUT-GLOP的标准格式是888-4567，310-GINO的标准格式是310-4466，3-10-10-10的标准格式是310-1010。
如果两个号码有相同的标准格式，那么他们就是等同的（相同的拨号）
你的公司正在为本地的公司编写一个电话号码薄。作为质量控制的一部分，你想要检查是否有两个和多个公司拥有相同的电话号码。
Input
Output
思路：把所有电话号码转成int型整数，再排序。计数判断重复的，输出。
UVa：
/*0.209s*/
#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
using namespace std;
const int trans[] = {0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9};
int ar[100010];
int main()
{
int t, n, i, j;
char ch;
bool fg;
scanf("%d", &t);
while (t--)
{
memset(ar, 0, sizeof(ar));
scanf("%d\n", &n);
for (i = 0; i < n; ++i)
while ((ch = getchar()) != 10)
if (ch != '-') ar[i] = ar[i] * 10 + (isdigit(ch) ? ch & 15 : trans[ch & 31]);
sort(ar, ar + n);
fg = true;
for (i = 0; i < n; i = j)
{
for (j = i; j < n && ar[j] == ar[i]; ++j)
;
if (j - i > 1) printf("%03d-%04d %d\n", ar[i] / 10000, ar[i] % 10000, j - i), fg = false;
}
if (fg) puts("No duplicates.");
if (t) putchar(10);
}
return 0;
}
POJ：
/*266ms,748KB*/
#include<cstdio>
#include<cctype>
#include<algorithm>
using namespace std;
const int trans[] = {0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9};
int ar[100010];
int main()
{
int n, i, j;
char ch;
bool fg;
scanf("%d\n", &n);
for (i = 0; i < n; ++i)
while ((ch = getchar()) != 10)
if (ch != '-') ar[i] = ar[i] * 10 + (isdigit(ch) ? ch & 15 : trans[ch & 31]);
sort(ar, ar + n);
for (i = 0; i < n; i = j)
{
for (j = i; j < n && ar[j] == ar[i]; ++j)
;
if (j - i > 1) printf("%03d-%04d %d\n", ar[i] / 10000, ar[i] % 10000, j - i), fg = false;
}
if (fg) puts("No duplicates.");
return 0;
}