POJ 1002 487-3279

来源：互联网发布：java微信支付回调demo 编辑：程序博客网时间：2024/06/05 21:52

这道题的思路是：

首先利用char map[] = "2223334445556667-77888999-"; //ABCDEFGHIJKLMNOPQRSTUVWXYZ 对数据进行转换，转换之后的数据只包含-和数字，然后去掉字符串中的-，并重新添加上‘-’。最后排序，统计字符串的数据。

#include <iostream>#include <cstdio>#include <vector>#include <string>#include <functional>#include <algorithm>using namespace std;char map[] = "2223334445556667#77888999#";                  //ABCDEFGHIJKLMNOPQRSTUVWXYZvoid visited(char &ch) // visit and format strings{   if (ch >= 'A' && ch <= 'Z')    ch=map[ch-'A']; // ch equals to its real number}void replace_m( string &s){    string::iterator it = s.begin();    for (; it != s.end(); it ++)    {        if (*it <='P' && *it >= 'A')        {            *it = '2'+(*it-'A')/3;        }        else if(*it >'P' && *it < 'Z')        {            *it = '7'+(*it-'Q')/3;        }    }}int main (){    //freopen("1.txt","r",stdin);    int num = 0;    cin >> num;    vector<string> svec;    int cts = 0;    string inpts;    {        while (num --)        {            cin >> inpts;            inpts.erase(remove_if(inpts.begin(),inpts.end(),bind2nd(equal_to<char>(),'-')),inpts.end());            inpts.erase(remove_if(inpts.begin(),inpts.end(),bind2nd(equal_to<char>(),'Q')),inpts.end());            inpts.erase(remove_if(inpts.begin(),inpts.end(),bind2nd(equal_to<char>(),'Z')),inpts.end());            replace_m(inpts);            svec.push_back(inpts);        }        sort (svec.begin(),svec.end());        string ss = *(svec.begin());        vector<string>::iterator it = svec.begin();        ++it;        int sct = 1;        int flag = 0;        int tamp = 0;        for (; it != svec.end(); it ++)        {            if (!ss.compare(*it))            {                sct ++;                tamp = 1;            }            else            {                if(sct >=2)                {                    ss.insert(3,"-");                    flag = 1;                    cout << ss <<' ' << sct <<endl;                    tamp = 0;                }                sct = 1;                ss = *it;            }        }        if(tamp && sct>=2)        {            ss.insert(3,"-");            flag = 1;            cout << ss <<' ' << sct <<endl;            tamp = 0;        }        if(!flag) cout<<"No duplicates."<<endl;    }}

这道题木在最后输出的时候，由于统计的时候考虑疏忽，当num=2的时候，出现错误。