uva 755(排序、检索)

题目：

Businesses like to have memorable telephone numbers. One way to make a telephone number memorableis to have it spell a memorable word or phrase. For example, you can call the University of Waterloo bydialing the memorable TUT-GLOP. Sometimes only part of the number is used to spell a word. Whenyou get back to your hotel tonight you can order a pizza from Gino's by dialing 310-GINO. Anotherway to make a telephone number memorable is to group the digits in a memorable way. You couldorder 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 andfourth 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 thestandard 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 thequality 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 numbersin the directory (up to 100,000) as a positive integer alone on the line. The remaining lines list thetelephone numbers in the directory, with each number alone on a line. Each telephone number consistsof a string composed of decimal digits, uppercase letters (excluding Q and Z) and hyphens. Exactlyseven 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. Theline should give the telephone number in standard form, followed by a space, followed by the numberof times the telephone number appears in the directory. Arrange the output lines by telephone numberin ascending lexicographical order. If there are no duplicates in the input print the line:

No duplicates.

Print a blank line between datasets.

sample intput

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

题解：将字符串中的字母按题目给的映射转成数字，按字典序排序后按标准的电话格式输出，并输出重复的次数，1次的不输出，记得位数不够要前补零。刚开始用string来存储号码不停的超时，int方式存储AC了。。

#include <iostream> #include <cstdio> #include <string>#include <cstring>#include <algorithm>using namespace std;const int MAX = 100010;string str[MAX];int ans[MAX];const int judge[24] = {2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9};int cmp(const void* s1, const void* s2) {return *(int *)s1 > *(int *)s2;}int main() {int cases, t, n;cin >> cases;while (cases--) {cin >> t;getchar();int k;for (int i = 0; i < t; i++) {getline(cin, str[i]);int len = str[i].size();k = 1000000;for (int j = 0; j < len; j++) {if (str[i][j] >= 'A' && str[i][j] <= 'P') {ans[i] += judge[str[i][j] - 'A'] * k;k /= 10;}else if (str[i][j] >= 'R' && str[i][j] <= 'Z') {ans[i] += judge[str[i][j] - 'A' - 1] * k;k /= 10;}else if (str[i][j] >= '0' && str[i][j] <= '9') {ans[i] += (str[i][j] - '0') * k;k /= 10;}}}qsort(ans, t, sizeof(ans[0]), cmp);int flag = 1, flag1 = 0;for (int i = 0; i < t - 1; i++) {if (ans[i] == ans[i + 1])flag++;else if (flag > 1) {printf("%03d-%04d %d\n", ans[i] / 10000, ans[i] % 10000, flag);//不足八位要补零！！flag1++;flag = 1;} }if (ans[t - 1] == ans[t - 2]) {printf("%03d-%04d %d\n", ans[t - 1] / 10000, ans[t - 1] % 10000, flag);flag1++;}if (flag1 == 0)cout << "No duplicates." << endl;if (cases)cout << endl;memset(ans, 0, sizeof(ans));}return 0;}