ZOJ_3501 Roman Order

　　这题应该是个水题，可是我比题还水。。。TT，用map写的，存对应的roma数字时存错啦。。。正好在比赛，我晕，WA了两次，我的排名啊。。。TT

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <map>

using namespace std;

map<int, string> m;

struct roma
{
    int x;
    string s;
}roma[10010];

void init()
{
    m[1] = "I";m[2] = "II";m[3] = "III";m[4] = "IV";
    m[5] = "V";m[6] = "VI";m[7] = "VII";m[8] = "VIII";
    m[9] = "IX";m[10] = "X";m[20] = "XX";m[30] = "XXX";
    m[40] = "XL";m[50] = "L"; m[60] = "LX"; m[70] = "LXX";m[80] = "LXXX";
    m[90] = "XC";m[100] = "C"; m[200] = "CC";m[300] = "CCC";
    m[400] = "CD"; m[500] = "D"; m[600] = "DC"; m[700] = "DCC";
    m[800] = "DCCC"; m[900] = "CM"; m[1000] = "M"; m[2000] = "MM";
    m[3000] = "MMM";
}

int cmp ( const void * a , const void * b )
{
    struct roma *c = (struct roma *)a;
    struct roma *d = (struct roma *)b;
    return c->s > d->s;
}

int main()
{
    //freopen("data.in", "r", stdin);

    int t, n, i, f, tmp;
    init();
    string str;

    while(scanf("%d", &t) != EOF)
    {
        while(t--)
        {
            scanf("%d", &n);
            for(i = 0; i < n; i++)
            {
                scanf("%d", &f);
                tmp = f;
                str.clear();
                if(f/1000)
                {
                    str += m[(f/1000)*1000];
                    f %= 1000;
                }
                if(f/100)
                {
                    str += m[(f/100)*100];
                    f %= 100;
                }
                if(f/10)
                {
                    str += m[(f/10)*10];
                    f %= 10;
                }
                if(f)
                {
                    str += m[f];
                }
                roma[i].x = tmp;
                roma[i].s.clear();
                roma[i].s = str;
            }

            qsort(roma, n, sizeof(roma[0]), cmp);

            for(i = 0; i < n-1; i++)
            {
                printf("%d ", roma[i].x);
                //cout << roma[i].s << " " ;
            }
            printf("%d\n", roma[n-1].x);
            //cout << roma[i].s << endl;
        }
    }
    return 0;
}