题目传送门
1 /*
2 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！
3 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
4
5 注意：1. Both A and B will have same number of digits 两个数字位数相同
6 2. which is no larger than 10^6 不是大小，而是长度不超过1e6
7 */
8 #include <cstdio>
9 #include <iostream>
10 #include <algorithm>
11 #include <cstring>
12 #include <string>
13 #include <cmath>
14 using namespace std;
15
16 const int MAXN = 1e6 + 10;
17 const int INF = 0x3f3f3f3f;
18 char s1[MAXN], s2[MAXN];
19
20 int main(void) //HDOJ 4726 Kia's Calculation
21 {
22 //freopen ("K.in", "r", stdin);
23
24 int t, cas = 0;
25 int cnt1[11], cnt2[11], cnt3[11];
26
27 scanf ("%d", &t);
28 while (t--)
29 {
30 scanf ("%s", &s1);
31 scanf ("%s", &s2);
32
33 printf ("Case #%d: ", ++cas);
34
35 int len = strlen (s1);
36 if (strcmp (s1, "0") == 0)
37 {
38 printf ("%s
", s2); continue;
39 }
40 else if (strcmp (s2, "0") == 0)
41 {
42 printf ("%s
", s1); continue;
43 }
44
45 memset (cnt1, 0, sizeof (cnt1));
46 memset (cnt2, 0, sizeof (cnt2));
47 memset (cnt3, 0, sizeof (cnt3));
48
49 for (int i=0; i<len; ++i)
50 {
51 cnt1[s1[i]-'0']++; cnt2[s2[i]-'0']++;
52 }
53
54 int ii = 1, jj = 1, mx = -1;
55 for (int i=1; i<=9; ++i)
56 {
57 if (cnt1[i] == 0) continue;
58 for (int j=1; j<=9; ++j)
59 {
60 if (cnt2[j] == 0) continue;
61 int tmp = (i + j) % 10;
62 if (tmp > mx)
63 {
64 mx = tmp; ii = i; jj = j;
65 }
66 }
67 }
68 cnt1[ii]--; cnt2[jj]--;
69 if (!mx)
70 {
71 puts ("0"); continue;
72 }
73
74 for (int i=9; i>=0; --i)
75 {
76 for (int j=0; j<=9; ++j)
77 {
78 for (int k=0; k<=9; ++k)
79 {
80 if ((j+k)%10==i && cnt1[j] && cnt2[k])
81 {
82 int tmp = min (cnt1[j], cnt2[k]);
83 cnt1[j] -= tmp; cnt2[k] -= tmp;
84 cnt3[i] += tmp;
85 }
86 }
87 }
88 }
89
90 printf ("%d", mx);
91 for (int i=9; i>=0; --i)
92 {
93 for (int j=0; j<cnt3[i]; ++j) printf ("%d", i);
94 }
95 puts ("");
96 }
97
98
99 return 0;
100 }