贪心 HDOJ 4726 Kia's Calculation

题目传送门

 /*
     这题交给队友做，做了一个多小时，全排列，RE数组越界，赛后发现读题读错了，囧！
     贪心：先确定最高位的数字，然后用贪心的方法，越高位数字越大

     注意：1. Both A and B will have same number of digits 两个数字位数相同
             2. which is no larger than 10^6 不是大小，而是长度不超过1e6
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <string>
 #include <cmath>
 using namespace std;

 const int MAXN = 1e6 + ;
 const int INF = 0x3f3f3f3f;
 char s1[MAXN], s2[MAXN];

 int main(void)        //HDOJ 4726    Kia's Calculation
 {
     //freopen ("K.in", "r", stdin);

     int t, cas = ;
     int cnt1[], cnt2[], cnt3[];

     scanf ("%d", &t);
     while (t--)
     {
         scanf ("%s", &s1);
         scanf ("%s", &s2);

         printf ("Case #%d: ", ++cas);

         int len = strlen (s1);
         if (strcmp (s1, "") == )
         {
             printf ("%s\n", s2);    continue;
         }
         else if (strcmp (s2, "") == )
         {
             printf ("%s\n", s1);    continue;
         }

         memset (cnt1, , sizeof (cnt1));
         memset (cnt2, , sizeof (cnt2));
         memset (cnt3, , sizeof (cnt3));

         for (int i=; i<len; ++i)
         {
             cnt1[s1[i]-'']++;    cnt2[s2[i]-'']++;
         }

         int ii = , jj = , mx = -;
         for (int i=; i<=; ++i)
         {
             if (cnt1[i] == )    continue;
             for (int j=; j<=; ++j)
             {
                 if (cnt2[j] == )    continue;
                 int tmp = (i + j) % ;
                 if (tmp > mx)
                 {
                     mx = tmp;    ii = i;    jj = j;
                 }
             }
         }
         cnt1[ii]--;    cnt2[jj]--;
         if (!mx)
         {
             puts ("");        continue;
         }

         for (int i=; i>=; --i)
         {
             for (int j=; j<=; ++j)
             {
                 for (int k=; k<=; ++k)
                 {
                     if ((j+k)%==i && cnt1[j] && cnt2[k])
                     {
                         int tmp = min (cnt1[j], cnt2[k]);
                         cnt1[j] -= tmp;    cnt2[k] -= tmp;
                         cnt3[i] += tmp;
                     }
                 }
             }
         }

         printf ("%d", mx);
         for (int i=; i>=; --i)
         {
             for (int j=; j<cnt3[i]; ++j)    printf ("%d", i);
         }
         puts ("");
     }

     return ;
 }