Professor GukiZ doesn’t accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Sample test(s)
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
题目大意：给你三个字符串s1, s2, s3。现在可以交换字符串s1中的字符，要求最终在字符串s1中找到的s2, s3字符串数量最多。
解题思路：先枚举其中一个串在主串中出现的次数，然后算出另一个串在主串中的次数，求次数和的最大。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int N = 100005;
const int M = 30;
char s[3][N];
int vis[3][M];
int ans, a, b;
int main() {
for (int i = 0; i < 3; i++) {
scanf("%s", s[i]);
int l = strlen(s[i]);
for (int j = 0; j < l; j++) {
vis[i][s[i][j] - 'a']++;
}
}
ans = 0;
for (int i = 0; ; i++) {
int flag = N;
for (int j = 0; j < 26; j++) {
if (vis[0][j] < vis[1][j] * i) {
flag = 0;
break;
}
}
if (!flag) break;
int p = N;
for (int j = 0; j < 26; j++) {
if (vis[2][j]) {
p = min(p, (vis[0][j] - vis[1][j] * i) / vis[2][j]);
}
}
if (i + p > ans) {
ans = i + p;
a = i;
b = p;
}
}
for (int i = 0; i < a; i++) {
printf("%s", s[1]);
}
for (int i = 0; i < b; i++) {
printf("%s", s[2]);
}
for (int i = 0; i < 26; i++) {
for (int j = 0; j < vis[0][i] - vis[1][i] * a - vis[2][i] * b; j++) {
printf("%c", i + 'a');
}
}puts("");
return 0;
}
