B. ZgukistringZ
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/551/problem/B
Description
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
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 Input
aaa
a
b
Sample Output
aaa
HINT
题意
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
题解：
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
代码：
#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=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** int num[];
int tmp[];
int num11[];
int num22[];
int main()
{
//test;
string a,b,c;
cin>>a>>b>>c;
for(int i=;i<a.size();i++)
num[a[i]-'a']++;
for(int i=;i<b.size();i++)
num11[b[i]-'a']++;
for(int i=;i<c.size();i++)
num22[c[i]-'a']++;
int flag=;
int ans1=,ans2=,ans3=;
int anss=;
for(int i=;i<=a.size();i++)
{
int flag=;
for(int j=;j<;j++)
tmp[j]=num[j];
for(int j=;j<;j++)
{
if(num11[j]*i>tmp[j])
flag=;
else
tmp[j]-=num11[j]*i;
}
if(flag==)
anss++;
if(anss==)
break;
int flag2=inf;
for(int j=;j<;j++)
{
if(num22[j]>)
flag2=min(flag2,tmp[j]/num22[j]);
}
if(flag2==inf)
flag2=;
if(flag)
flag2+=i;
if(ans3<flag2)
{
ans1=i;
ans2=flag2-i;
ans3=flag2;
}
}
for(int i=;i<ans1;i++)
cout<<b;
for(int i=;i<ans2;i++)
cout<<c;
for(int i=;i<;i++)
num[i]=num[i]-num11[i]*ans1;
for(int i=;i<;i++)
num[i]=num[i]-num22[i]*ans2;
for(int i=;i<;i++)
{
while(num[i]>)
{
num[i]--;
printf("%c",i+'a');
}
}
}