array(2) { ["docs"]=> array(10) { [0]=> array(10) { ["id"]=> string(3) "428" ["text"]=> string(77) "Visual Studio 2017 单独启动MSDN帮助(Microsoft Help Viewer)的方法" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(8) "DonetRen" ["tagsname"]=> string(55) "Visual Studio 2017|MSDN帮助|C#程序|.NET|Help Viewer" ["tagsid"]=> string(23) "[401,402,403,"300",404]" ["catesname"]=> string(0) "" ["catesid"]=> string(2) "[]" ["createtime"]=> string(10) "1511400964" ["_id"]=> string(3) "428" } [1]=> array(10) { ["id"]=> string(3) "427" ["text"]=> string(42) "npm -v;报错 cannot find module "wrapp"" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(4) "zzty" ["tagsname"]=> string(50) "node.js|npm|cannot find module "wrapp“|node" ["tagsid"]=> string(19) "[398,"239",399,400]" ["catesname"]=> string(0) "" ["catesid"]=> string(2) "[]" ["createtime"]=> string(10) "1511400760" ["_id"]=> string(3) "427" } [2]=> array(10) { ["id"]=> string(3) "426" ["text"]=> string(54) "说说css中pt、px、em、rem都扮演了什么角色" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(12) "zhengqiaoyin" ["tagsname"]=> string(0) "" ["tagsid"]=> string(2) "[]" ["catesname"]=> string(0) "" ["catesid"]=> string(2) "[]" ["createtime"]=> string(10) "1511400640" ["_id"]=> string(3) "426" } [3]=> array(10) { ["id"]=> string(3) "425" ["text"]=> string(83) "深入学习JS执行--创建执行上下文(变量对象,作用域链,this)" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(7) "Ry-yuan" ["tagsname"]=> string(33) "Javascript|Javascript执行过程" ["tagsid"]=> string(13) "["169","191"]" ["catesname"]=> string(0) "" ["catesid"]=> string(2) "[]" ["createtime"]=> string(10) "1511399901" ["_id"]=> string(3) "425" } [4]=> array(10) { ["id"]=> string(3) "424" ["text"]=> string(30) "C# 排序技术研究与对比" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(9) "vveiliang" ["tagsname"]=> string(0) "" ["tagsid"]=> string(2) "[]" ["catesname"]=> string(8) ".Net Dev" ["catesid"]=> string(5) "[199]" ["createtime"]=> string(10) "1511399150" ["_id"]=> string(3) "424" } [5]=> array(10) { ["id"]=> string(3) "423" ["text"]=> string(72) "【算法】小白的算法笔记:快速排序算法的编码和优化" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(9) "penghuwan" ["tagsname"]=> string(6) "算法" ["tagsid"]=> string(7) "["344"]" ["catesname"]=> string(0) "" ["catesid"]=> string(2) "[]" ["createtime"]=> string(10) "1511398109" ["_id"]=> string(3) "423" } [6]=> array(10) { ["id"]=> string(3) "422" ["text"]=> string(64) "JavaScript数据可视化编程学习(二)Flotr2,雷达图" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(7) "chengxs" ["tagsname"]=> string(28) "数据可视化|前端学习" ["tagsid"]=> string(9) "[396,397]" ["catesname"]=> string(18) "前端基本知识" ["catesid"]=> string(5) "[198]" ["createtime"]=> string(10) "1511397800" ["_id"]=> string(3) "422" } [7]=> array(10) { ["id"]=> string(3) "421" ["text"]=> string(36) "C#表达式目录树(Expression)" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(4) "wwym" ["tagsname"]=> string(0) "" ["tagsid"]=> string(2) "[]" ["catesname"]=> string(4) ".NET" ["catesid"]=> string(7) "["119"]" ["createtime"]=> string(10) "1511397474" ["_id"]=> string(3) "421" } [8]=> array(10) { ["id"]=> string(3) "420" ["text"]=> string(47) "数据结构 队列_队列实例:事件处理" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(7) "idreamo" ["tagsname"]=> string(40) "C语言|数据结构|队列|事件处理" ["tagsid"]=> string(23) "["246","247","248",395]" ["catesname"]=> string(12) "数据结构" ["catesid"]=> string(7) "["133"]" ["createtime"]=> string(10) "1511397279" ["_id"]=> string(3) "420" } [9]=> array(10) { ["id"]=> string(3) "419" ["text"]=> string(47) "久等了,博客园官方Android客户端发布" ["intro"]=> string(288) "目录 ECharts 异步加载 ECharts 数据可视化在过去几年中取得了巨大进展。开发人员对可视化产品的期望不再是简单的图表创建工具,而是在交互、性能、数据处理等方面有更高的要求。 chart.setOption({ color: [ " ["username"]=> string(3) "cmt" ["tagsname"]=> string(0) "" ["tagsid"]=> string(2) "[]" ["catesname"]=> string(0) "" ["catesid"]=> string(2) "[]" ["createtime"]=> string(10) "1511396549" ["_id"]=> string(3) "419" } } ["count"]=> int(200) } 222 POJ 1002 487-3279 - 爱码网
A - 487-3279
Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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

 

 
 1 //hash直接过
 2 #include<cstdio>
 3 #include<cstring>
 4 int hash[10010000]={0};
 5 int zhuanzhi(int n)
 6 {
 7     int sum=1,i;
 8     for(i=1;i<n;i++)
 9         sum*=10;
10     return sum;
11 }
12 int zhuanhuan(char f[])
13 {
14     int sum=0,i;
15     int t=strlen(f);
16     for(i=t-1;i>=0;i--)
17     {
18         sum=sum+(f[i]-'0')*zhuanzhi(t-i);
19     }
20     return sum;
21 }
22 int main()
23 {
24     int zong,i,j;
25     scanf("%d",&zong);
26     while(zong--)
27     {
28         char f[100],g[100];
29         scanf("%s",f);
30         int s=-1;
31         for(i=0;f[i]!='\0';i++)
32         {
33             switch(f[i])
34             {
35             case '0':
36                 g[++s]='0';
37                 break;
38             case '1':
39                 g[++s]='1';
40                 break;
41             case 'A':case 'B':case 'C':case '2':
42                 g[++s]='2';
43                 break;
44             case 'D':case 'E':case 'F':case '3':
45                 g[++s]='3';
46                 break;
47             case 'G':case 'H':case 'I':case '4':
48                 g[++s]='4';
49                 break;
50             case 'J':case 'K':case 'L':case '5':
51                 g[++s]='5';
52                 break;
53             case 'M':case 'N':case 'O':case '6':
54                 g[++s]='6';
55                 break;
56             case 'P':case 'R':case 'S':case '7':
57                 g[++s]='7';
58                 break;
59             case 'T':case 'U':case 'V':case '8':
60                 g[++s]='8';
61                 break;
62             case 'W':case 'X':case 'Y':case '9':
63                 g[++s]='9';
64                 break;
65             }
66         }
67         g[++s]='\0';
68         hash[zhuanhuan(g)]++;
69     }
70     int count=0;
71     for(i=0;i<10000000;i++)
72         if(hash[i]>1)
73         {
74             printf("%03d-%04d %d\n",i/10000,i%10000,hash[i]);
75             count++;
76         }
77     if(count==0)printf("No duplicates.\n");
78     return 0;
79 }

 

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