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

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最新推荐文章于 2025-09-12 17:31:31 发布

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本文提供了一种方法来解决特定的字符串匹配问题，目标是在给定的字符串a中通过交换，使得b和c串在a中出现的非重叠子串数量尽可能多。该方法涉及暴力枚举和贪心策略，旨在找到最优解。通过实例演示了如何实现这一过程，并提供了代码示例。

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?

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 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=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');
        }
    }
}