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
?
The first line contains string
a
, the second line contains string
b
, and the third line contains string
c
(
1 ≤ |
a
|, |
b
|, |
c
| ≤ 10
5
, 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.
Find one of possible strings
k
, as described in the problem statement. If there are multiple possible answers, print any of them.
aaa a b
aaa
pozdravstaklenidodiri niste dobri
nisteaadddiiklooprrvz
abbbaaccca ab aca
ababacabcc
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
.
题意：
给出一个3个串s1,s2,s3要求改变s1串字母的位置,使得s2,s3在在改变后的串出现的总次数尽量多，而且不算重叠部分
思路：
先枚举其中一串出现的次数，然后计算不同情况下另外一串能出现的最大次数，最后看总次数最大的情况
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <list>
#include <algorithm>
#include <climits>
using namespace std;
#define lson 2*i
#define rson 2*i+1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 100005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
const int mod = 1e9+7;
char s1[N],s2[N],s3[N];
int len1,len2,len3;
int h1[30],h2[30],h3[30];
int cnt2,cnt3,cnt;
int main()
{
int i,j,k;
while(~scanf("%s%s%s",s1,s2,s3))
{
len1 = strlen(s1);
len2 = strlen(s2);
len3 = strlen(s3);
MEM(h1,0);
MEM(h2,0);
MEM(h3,0);
for(i = 0; i<len1; i++)
h1[s1[i]-'a']++;
for(i = 0; i<len2; i++)
h2[s2[i]-'a']++;
for(i = 0; i<len3; i++)
h3[s3[i]-'a']++;
cnt = 0;
for(i=1;; i++)
{
int p = N;
for(j = 0; j<26; j++)
{
if(h1[j]-h2[j]*i<0)
{
p=0;
break;
}
}
if(!p) break;
int p2 = N;
for(j = 0; j<26; j++)
{
if(h3[j])
{
p2 = min(p2,(h1[j]-h2[j]*i)/h3[j]);
}
}
if(i+p2>cnt)
{
cnt = i+p2;
cnt2 = i;
cnt3 = p2;
}
}
for(i=1;; i++)
{
int p = N;
for(j = 0; j<26; j++)
{
if(h1[j]-h3[j]*i<0)
{
p=0;
break;
}
}
if(!p) break;
int p2 = N;
for(j = 0; j<26; j++)
{
if(h2[j])
{
p2 = min(p2,(h1[j]-h3[j]*i)/h2[j]);
}
}
if(i+p2>cnt)
{
cnt = i+p2;
cnt3 = i;
cnt2 = p2;
}
}
for(i = 0; i<cnt2; i++)
printf("%s",s2);
for(i = 0; i<cnt3; i++)
printf("%s",s3);
for(i = 0; i<26; i++)
{
for(j = 0; j<h1[i]-cnt2*h2[i]-cnt3*h3[i]; j++)
printf("%c",i+'a');
}
printf("\n");
}
return 0;
}