Educational Codeforces Round 25 D. Suitable Replacement

最新推荐文章于 2017-08-22 22:09:20 发布

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最新推荐文章于 2017-08-22 22:09:20 发布

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分类专栏： codeforce 傻逼题

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本文介绍了一种字符串替换问题的解决方法，旨在通过特定的算法来最大化目标字符串与模板字符串的匹配次数。具体而言，对于含有特殊字符“?”的输入字符串s，通过对这些特殊字符进行最优替换，使得s中能包含尽可能多的模板字符串t。文章提供了详细的实现思路及代码示例。

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You are given two strings s and t consisting of small Latin letters, string s can also contain '?' characters.

Suitability of string s is calculated by following metric:

Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings s, you choose the one with the largest number of non-intersecting occurrences of string t. Suitability is this number of occurrences.

You should replace all '?' characters with small Latin letters in such a way that the suitability of string s is maximal.

Input

The first line contains string s (1 ≤ |s| ≤ 106).

The second line contains string t (1 ≤ |t| ≤ 106).

Output

Print string s with '?' replaced with small Latin letters in such a way that suitability of that string is maximal.

If there are multiple strings with maximal suitability then print any of them.

   Examples
 
     input
   
?aa?
ab

     output
   
baab

     input
   
??b?
za

     output
   
azbz

     input
   
abcd
abacaba

     output
   
abcd

Note

In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2.

In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them.

In the third example there are no '?' characters and the suitability of the string is 0.

题意：给你一个长度为1e6串，下面给一个模版串，问你输出最大使用下面的模版的串。?是可以替代为任意字符的地方。

做法：看代码。

//china no.1
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
using namespace std;

#define pi acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x) memset(x,0,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
//const int dx[]={-1,0,1,0,-1,-1,1,1};
//const int dy[]={0,1,0,-1,1,-1,1,-1};
const int maxn=1e3+5;
const int maxx=1e6+100;
const double EPS=1e-7;
const int MOD=10000007;
#define mod(x) ((x)%MOD);
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while

inline int Scan()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')flag=1;
    else if(ch>='0' && ch<='9')res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';
    return flag ? -res : res;
}

int main()
{
	string s,t;
	int i,j,hash[26]={0};

	cin>>s>>t;
	for(i=0;i<s.size();i++) hash[s[i]-97]++;//把s[i]的每个字符都记录下来。

	int z=0;

	for(i=0;i<s.size();i++)
    {

		if(s[i]=='?')//当然?里面应该是把t串的字符从头到尾的用一遍这个串就至少有了一个t串
            //但有可能本来s串里有t串的部分字母。所以刚开始要存一个s串字符的多少的数组
		{
			z++;
			z%=t.size();

			if(hash[t[z]-97] > 0)//如果t[z]这个位置s串中有这个字符就跳过
			{
				hash[t[z]-97]--;//就把这个字符--
				i--;//回到上一个位置 下一个位置还是?
			}
			else
				s[i]=t[z];
		}
	}
	cout<<s<<endl;
}