Codeforces Round #307 (Div. 2) B. ZgukistringZ 暴力

B. ZgukistringZ

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/551/problem/B

Description

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?

[b]Input[/b]

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.
[b]Output[/b]

Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.

[b]Sample Input[/b]

aaa
a
b

[b]Sample Output[/b]

aaa

HINT

[b]题意[/b]

给你a,b,c三个串，让你随意交换a串的位置，让b串和c串在a串里面不重复的出现最多次

[b]题解：[/b]

B题，就老老实实想暴力就好，直接暴力枚举b串出现的次数，然后再算出c串出现的最多次数，然后搞一搞就好了

蛤蛤

[b]代码：[/b]

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
//**************************************************************************************

int num[30];
int tmp[30];
int num11[30];
int num22[30];
int main()
{
//test;
string a,b,c;
cin>>a>>b>>c;
for(int i=0;i<a.size();i++)
num[a[i]-'a']++;
for(int i=0;i<b.size();i++)
num11[b[i]-'a']++;
for(int i=0;i<c.size();i++)
num22[c[i]-'a']++;
int flag=1;
int ans1=0,ans2=0,ans3=0;
int anss=0;
for(int i=0;i<=a.size();i++)
{
int flag=1;
for(int j=0;j<30;j++)
tmp[j]=num[j];
for(int j=0;j<30;j++)
{
if(num11[j]*i>tmp[j])
flag=0;
else
tmp[j]-=num11[j]*i;
}
if(flag==0)
anss++;
if(anss==5)
break;
int flag2=inf;
for(int j=0;j<30;j++)
{
if(num22[j]>0)
flag2=min(flag2,tmp[j]/num22[j]);
}
if(flag2==inf)
flag2=0;
if(flag)
flag2+=i;
if(ans3<flag2)
{
ans1=i;
ans2=flag2-i;
ans3=flag2;
}
}
for(int i=0;i<ans1;i++)
cout<<b;
for(int i=0;i<ans2;i++)
cout<<c;
for(int i=0;i<30;i++)
num[i]=num[i]-num11[i]*ans1;
for(int i=0;i<30;i++)
num[i]=num[i]-num22[i]*ans2;
for(int i=0;i<30;i++)
{
while(num[i]>0)
{
num[i]--;
printf("%c",i+'a');
}
}
}