UVA 755 - 487--3279

       给出一些字符串，每个字符串对应一个电话号码，要求输出所有字符串中所包含的每种电话号码的个数（只输出出现次数大于1的），如果直接搜索的话会TLE，因为总共电话号码为100000，可以先把所有的字符串按题目要求转化成数字，然后用一个数组hash[10000000]存下所有的电话号码，如果输入的字符串转化为数字为a，则hash[a]++;

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<iostream>
using namespace std;
char tel[100001][100];
int hash[10000000];
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        memset(hash,0,sizeof(hash));
        int num,te=0;;
        cin>>num;
        char c;
        cin.get(c);
        int temp[100];
        for(int i=0; i<num; i++)
        {
            cin.getline(tel[i],100,\'\\n\');
            int len=strlen(tel[i]);
            int k=0;
            for(int j=0; j<len; j++)
            {
                if(tel[i][j]>=\'A\'&&tel[i][j]<=\'C\')
                    temp[k++]=2;
                else if(tel[i][j]>=\'D\'&&tel[i][j]<=\'F\')
                    temp[k++]=3;
                else if(tel[i][j]>=\'G\'&&tel[i][j]<=\'I\')
                    temp[k++]=4;
                else if(tel[i][j]>=\'J\'&&tel[i][j]<=\'L\')
                    temp[k++]=5;
                else if(tel[i][j]>=\'M\'&&tel[i][j]<=\'O\')
                    temp[k++]=6;
                else if(tel[i][j]>=\'P\'&&tel[i][j]<=\'S\')
                    temp[k++]=7;
                else if(tel[i][j]>=\'T\'&&tel[i][j]<=\'V\')
                    temp[k++]=8;
                else if(tel[i][j]>=\'W\'&&tel[i][j]<=\'Y\')
                    temp[k++]=9;
                else if(tel[i][j]>=\'0\'&&tel[i][j]<=\'9\')
                    temp[k++]=tel[i][j]-\'0\';
            }
            int sum=0;
            int ss=1000000;
            for(int j=0; j<7; j++)
            {
                sum+=temp[j]*ss;
                ss/=10;
            }
            hash[sum]++;
        }
        int flag=0;
        for(int i=0; i<10000000; i++)
        {
            if(hash[i]>1)
            {
                int a1=i/10000;
                int a2=i%10000;
                printf(\"%03d-%04d %d\\n\",a1,a2,hash[i]);

                flag=1;
            }
        }
        if(!flag)
        {
            cout<<\"No duplicates.\"<<endl;
        }
        if(n)
        cout<<endl;
    }
    return 0;
}