题目传送门
1 /*
2 题意：任意排列第一个字符串，使得有最多的不覆盖a/b字符串出现
3 字符串处理/贪心：暴力找到最大能不覆盖的a字符串，然后在b字符串中动态得出最优解
4 恶心死我了，我最初想输出最多的a，再最多的b，然而并不能保证是最多的:(
5 */
6 #include <cstdio>
7 #include <cstring>
8 #include <string>
9 #include <iostream>
10 #include <algorithm>
11 #include <cmath>
12 using namespace std;
13
14 const int MAXN = 1e5 + 10;
15 const int INF = 0x3f3f3f3f;
16 char s[MAXN], a[MAXN], b[MAXN];
17 int cnt_s[30], cnt_a[30], cnt_b[30];
18 int len_s, len_a, len_b;
19
20 int main(void) //Codeforces Round #307 (Div. 2) B. ZgukistringZ
21 {
22 // freopen ("B.in", "r", stdin);
23
24 while (scanf ("%s", s) == 1)
25 {
26 scanf ("%s", a); scanf ("%s", b);
27 len_s = strlen (s); len_a = strlen (a); len_b = strlen (b);
28
29 memset (cnt_s, 0, sizeof (cnt_s));
30 memset (cnt_a, 0, sizeof (cnt_a));
31 memset (cnt_b, 0, sizeof (cnt_b));
32
33 for (int i=0; i<len_s; ++i) cnt_s[s[i]-'a']++;
34 for (int i=0; i<len_a; ++i) cnt_a[a[i]-'a']++;
35 for (int i=0; i<len_b; ++i) cnt_b[b[i]-'a']++;
36
37 int ans_a = 0, ans_b = 0; int tot = 100000;
38 for (int i=0; i<30; ++i) {if (cnt_a[i]) tot = min (tot, cnt_s[i] / cnt_a[i]);}
39 for (int i=0; i<=tot; ++i)
40 {
41 int p = 100000;
42 for (int j=0; j<30; ++j) {if (cnt_b[j]) p = min (p, (cnt_s[j] - i * cnt_a[j]) / cnt_b[j]);}
43 if (i + p > ans_a + ans_b) {ans_a = i; ans_b = p;}
44 }
45
46 for (int i=1; i<=ans_a; ++i) printf ("%s", a);
47 for (int j=1; j<=ans_b; ++j) printf ("%s", b);
48 for (int i=0; i<30; ++i)
49 {
50 for (int j=1; j<=cnt_s[i]-ans_a*cnt_a[i]-ans_b*cnt_b[i]; ++j)
51 printf ("%c", 'a' + i);
52 }
53 puts ("");
54 }
55
56 return 0;
57 }