codeforces825D-Suitable Replacement

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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.

题目大意：给你一个 s 字符串和一个 t 字符串,在 s 字符串中有一些？，然后我们需要使用一些字母来替换掉？，使得 s 中能有最多的字符串 t。

解题思路：我们需要使得 s 中包含尽可能多的 t ，所以我们肯定是用 t 中的字符来替换掉 s 中的？，所以我们在这里先存贮 s 中每个字符的数量，然后我们对于 s 中的每一个？进行替换，替换规则是从t0--无穷循环的t，看是否有，有的话就将 s 中的t0 字符减一，即表示用了，如果没有就将这个？变为当前的 t（i），然后再找下一个？，直到所有的？都被替换掉就可以了。

ac代码：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int maxn=1000006;
char s[maxn],t[maxn];
int num[300]; 
int main(){
	scanf("%s%s",&s,&t);
	int ls=strlen(s);
	int lt=strlen(t);
	for(int i=0;i<ls;i++){
		if(s[i]!='?')
		num[s[i]-'a']++;
	}
	int z=0;
	for(int i=0;i<ls;i++){
		if(s[i]=='?'){
			z++;
			z%=lt;
			if(num[t[z]-'a']>0){
				num[t[z]-'a']--;
				i--;
			}
			else{
				s[i]=t[z];
			}
		}
	}
	printf("%s",s);
	return 0;
}

题目链接：点击打开链接http://codeforces.com/problemset/problem/825/D