n组字母和最大

字母A-J，用0-9对应字母使得n组数据和最大，输入字符串前面保证非0
如输入组数据：
2
ABC
BCA
输出：
1875

思路：其实就是求和，对应字符乘以相应的量级，按系数排序
如上MAX(101A+110B+11C)
那B：9 A：8，C：7产生最大和
同时考虑类型为字符串涉及字符串加法

#include <algorithm>
#include <iostream>
#include <map>
#include <vector>
#include <string>
#include <math.h>
#define IMIN numeric_limits<int>::min()
#define IMAX numeric_limits<int>::max()
#define FR(i,n) for(int i=0;i<n;i++)
#define CLC(x) memset(x,0,sizeof(x))
#define FILL(x,c) memset(x,c,sizeof(x))
#define viter vector<int>::const_iterator
using namespace std;
string invert(string &src)
{
    string newStr=src;
    for(int i=src.length()-1,j=0; i>=0; --i,++j)
        newStr[j]=src[i];
    return newStr;
}
void inverse(string &s)
{
      for(int i=0;i<s.size()/2;++i){
          char tmp= s[i];
          s[i] = s[s.size()-1-i];
          s[s.size()-1-i] = tmp;
      }
}
string intAdd(string &rs1,string &rs2)
{
    string str1=invert(rs1);
    string str2=invert(rs2);
    if(str1.length()<str2.length())
        str1.swap(str2);

    for(size_t i=0; i!=str2.length(); ++i)
    {
        char c1=str1[i];
        char c2=str2[i];
        int t=((int)c1-48)+((int)c2-48);
        if(t>=10)
        {
            //进位
            int x=t/10;
            t%=10;
            size_t n=i+1;
            do
            {
                int t1=(int)str1[n]-48+x;
                if(t1>=10)
                {
                    str1[n]=(char)(t1%10+48);
                    ++n;
                }
                else
                {
                    str1[n]=(char)(t1+48);
                }
                if(n==str1.length())
                {
                    str1+="1";
                    break;
                }
                x=t1/10;
            }
            while(x!=0);

            str1[i]=(char)(t+48);
        }
        else
        {
            str1[i]=(char)(t+48);
        }
    }
    string &rstrResult=str1;
    string strOut=invert(rstrResult);
    return strOut;
}

void init_map(map<char,double> &wmap)   
{
    for(char i='A';i<'K';++i)wmap[i]=0.1;
}
string get_max(vector<string> &vstr)
{
    map<char,double> wmap;
    init_map(wmap);
    for(int i=0;i<vstr.size();++i)
    {
        double count =1.0;
        for(int j=0;j<vstr[i].size();++j)//0对应个位，逆序
        {

            wmap[vstr[i][j]]=wmap[vstr[i][j]]+count;
            count = count*10;
        }
    }
    map<double,char> nmap;
    map<char,double>::iterator it = wmap.begin();
    while(it!=wmap.end()){
        //cout<<"1="<<it->first<<" "<<it->second<<endl;
        if(nmap.count(1.0/double(it->second))){
            double dtmp=it->second-0.1;
            while(nmap.count(1.0/double(dtmp)))dtmp=dtmp-0.1;
            nmap[1.0/double(dtmp)]=it->first;
            //cout<<"SEC1: "<<1.0/double(dtmp)<<"fir1:"<<it->first<<endl;
        }
        else {
            //cout<<"SEC2: "<<1.0/double(it->second)<<" fir2:"<<it->first<<endl;
            nmap[1.0/double(it->second)]=it->first;
        }
        it++;
    }
    map<char,char> idmap;
    map<double,char>::iterator it0 = nmap.begin();
    char num='9';
    while(it0!=nmap.end()){
         idmap[it0->second]=char(num);
         cout<<idmap[it0->second]<<": "<<it0->second<<endl;
         num = num-1;
        it0++;
    }
    string res;   
    for(int i=0;i<vstr.size();++i)
    {
        inverse(vstr[i]);
        for(int j=0;j<vstr[i].size();++j)
        {
            vstr[i][j]=idmap[vstr[i][j]];   
        }
        cout<<vstr[i]<<endl;
        res=intAdd(res,vstr[i]);
    }
// inverse(res);
    return res;       
}

int main()
{

    int n;
    while(cin>>n)
    {
        vector<string> vstr;
        FR(i,n){
            string stmp;
            cin>>stmp;
            inverse(stmp);
            vstr.push_back(stmp);
        }
        cout<<get_max(vstr)<<endl;           
    }
    return 0;
}