POJ 之 1002 :487-3279
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 242418 | Accepted: 42978 |
Description
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
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
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 使用STL map做的,开的内存很大,不过这样写比较简单,清楚明了!
Accepted的代码 :
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <algorithm> using namespace std; int main()
{
int t;
map<string, int>ma;
map<string, int>::iterator it;
char s[1000];
char ch[30];
int i, len;
int sum;
while(scanf("%d", &t)!=EOF)
{
ma.clear();
sum=0;
while(t--)
{
scanf("%s", s);
len=strlen(s);
int e=0;
for(i=0; i<len; i++)
{
if(s[i]>='0' && s[i]<='9' )
{
ch[e++]=s[i];
if(e==3)
{
ch[e++]='-';
}
}
else if(s[i]>='A' && s[i]<='C' )
{
ch[e++]='2';
if(e==3)
{
ch[e++]='-';
}
}
else if(s[i]>='D' && s[i]<='F' )
{
ch[e++]= '3' ;
if(e==3)
{
ch[e++]='-';
}
}
else if(s[i]>='G' && s[i]<='I' )
{
ch[e++]='4';
if(e==3)
{
ch[e++]='-';
}
}
else if(s[i]>='J' && s[i]<='L' )
{
ch[e++] = '5';
if(e==3)
{
ch[e++]='-';
}
}
else if(s[i]>='M' && s[i]<='O' )
{
ch[e++]='6';
if(e==3)
{
ch[e++]='-';
}
}
else if(s[i]=='P' || s[i]=='R' ||s[i]=='S' )
{
ch[e++] = '7';
if(e==3)
{
ch[e++]='-';
}
}
else if(s[i]>='T' && s[i]<='V')
{
ch[e++]='8' ;
if(e==3)
{
ch[e++]='-';
}
}
else if(s[i]>='W' && s[i]<='Y' )
{
ch[e++]= '9';
if(e==3)
{
ch[e++]='-';
}
}
}
ch[e]='\0';
ma[ch]++;
}
for(it=ma.begin(); it!=ma.end(); it++)
{
if(it->second >1)
{
sum++;
printf("%s %d\n", it->first.c_str(), it->second );
}
}
if(sum==0)
{
printf("No duplicates.\n");
}
}
return 0;
}
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