POJ-1002

来源：http://poj.org/problem?id=1002&lang=zh-CN

这题用C++的string来写真的很蛋疼，效率低不说，输入输出真的麻烦。由此看出，C++的STL效率不行，不过除非是那种特别卡时的题，一般STL还是不错的。

这是C++的代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;

int dd[]={2,2,2,3,3,3,
    4,4,4,5,5,5,6,6,6,
    7,0,7,7,8,8,8,9,9,9,0};

string s[100005];
int f[100005]={0};
bool flag=true;

void print(string s,int x)
{
    s.insert(3,"-",1);  //这里用insert函数效率太低，不过我懒得写循环输出
    cout<<s<<" "<<x<<endl;
    flag=false;
}

int main()
{
    char ss[20];
    int n;
    scanf("%d\n",&n);   //这里一定要加换行符\n，不加肯定错
    for (int i=1;i<=n;i++)
    {
        s[i].clear();
        gets(ss);
        for (int j=0;ss[j];j++)
            if (ss[j]>='A' && ss[j]<='Z') s[i]+=dd[ss[j]-'A']+'0';
            else if (ss[j]!='-') s[i]+=ss[j];
    }
    sort(s+1,s+n+1);

    int x=1;
    
    string temp=s[1];
    for (int i=2;i<=n;i++)
    {
        if (s[i]==temp) x++;
        else {
            if (x>1) print(temp,x);
            x=1; temp=s[i];
        }
    }
    if (s[n]==s[n-1]) print(s[n],x);
    if (flag) printf("No duplicates.\n");
    return 0;
}

下面的是我早些时候写的pascal代码，效率还可以：

program eee;
const
  a:array['A'..'Z'] of char=('2','2','2','3','3','3','4','4','4','5','5','5',
                             '6','6','6','7','0','7','7','8','8','8','9','9','9',
                             '0');
var
n,i,j,k,l:longint;
d:array[0..100001] of string;
s:string;
flag:boolean;

procedure sort(l,r:longint);
var
i,j:longint;
k,x:string;
begin
  i:=l; j:=r; k:=d[(l+r) div 2];
  repeat
    while d[i]<k do inc(i);
    while d[j]>k do dec(j);
    if i<=j then
    begin
      x:=d[i]; d[i]:=d[j]; d[j]:=x;
      inc(i); dec(j);
    end;
  until i>j;
  if i<r then sort(i,r);
  if l<j then sort(l,j);
end;

begin
  readln(n);
  for i:=1 to n do
  begin
    readln(d[i]);
    s:='';
    for j:=1 to length(d[i]) do
    begin
      if d[i,j] in ['A'..'Z'] then s:=s+a[d[i,j]]
      else if d[i,j] in ['0'..'9'] then s:=s+d[i,j];
    end;
    insert('-',s,4);
    d[i]:=s;
  end;
  sort(1,n);
  flag:=true;
  s:=d[1]; k:=1;
  for i:=2 to n do
  begin
    if (d[i]<>s) and (k>1) then begin flag:=false; writeln(s,' ',k); s:=d[i]; k:=1; end
    else if d[i]<>s then s:=d[i]
    else inc(k);
  end;
  if d[n]=d[n-1] then begin writeln(d[n],' ',k); flag:=false; end;
  if flag then writeln('No duplicates.');
end.