Codeforces 825D Suitable Replacement【贪心】水题

D. Suitable Replacement

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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：

给你一个字符串S，和一个字符串T ，让你在字符串S中补全？使得在若干次交换S串之后得到的字符串，包含尽可能多的T串。

思路：

直接贪心搞即可，因为可以随便交换S串的字符位子，那么对应其实就是在问，补全?的情况下，S串能够最多拼出多少个T串。

那么统计T串各个字符出现的次数，然后S串补空缺即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char a[1500000];
char b[1500000];
int ned[500];
int have[500];
int main()
{
while(~scanf("%s%s",a,b))
{
memset(ned,0,sizeof(ned));
int lena=strlen(a);
int lenb=strlen(b);
for(int i=0;i<lenb;i++)
{
ned[b[i]]++;
}
for(int i=0;i<lena;i++)
{
have[a[i]]++;
}
int can=0x3f3f3f3f;
for(int i=0;i<300;i++)
{
if(ned[i]>0)
{
can=min(can,have[i]/ned[i]);
}
}
for(int i=0;i<300;i++)
{
if(ned[i]>0)
{
have[i]-=ned[i]*can;
}
}
int now=0;
for(int i=0;i<lena;i++)
{
if(a[i]=='?')
{
while(1)
{
int flag=0;
if(ned[now]==0)now++;
else
{
if(have[now]<ned[now])
{
have[now]++;
flag=1;
a[i]=now;
}
if(have[now]>=ned[now])
{
have[now]-=ned[now];
now++;
}
}
if(now>=300)now=0;
if(flag==1)break;
}
}
}
for(int i=0;i<lena;i++)
{
printf("%c",a[i]);
}
printf("\n");
}
}