codeforces825 D. Suitable Replacement 二分答案

最新推荐文章于 2019-08-07 10:25:00 发布

Dave_L

最新推荐文章于 2019-08-07 10:25:00 发布

阅读量459

点赞数

CC 4.0 BY-SA版权

分类专栏：暑期集训刷题算法复习（新手）二分

本文为博主原创文章，未经博主允许也可以转载。

本文链接：https://blog.youkuaiyun.com/FrankAx/article/details/77005334

暑期集训刷题算法复习（新手）同时被 2 个专栏收录

123 篇文章

订阅专栏

二分

33 篇文章

订阅专栏

本文介绍了一个字符串处理问题，目标是在给定的字符串S中通过替换问号字符来最大化包含目标子串T的数量。通过二分查找确定最优的替换方案，并详细解析了实现思路与代码实现细节。

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S ,T两个字符串，让你输出一个和S相同长度的串，并将S中的？改为小写字母，满足输出串中T串数量最多

思路：二分答案，模拟太烦人，搞得我乱死了（主要是太弱），先二分出最多有几个T串，然后输出的时候判断S串中哪个字母不够就输出，最后全够了就随便输出一个字母。

还有注意long long，我的一下午贡献给这题也是没谁了

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int AX = 1e6+66;
char a[AX];
char b[AX];
int numa[27];
int numb[27];
int num;
int lena,lenb;
bool check(int x){          //不用long long wa到死
	LL c=0;
	for(int i = 0 ; i < 26 ; i ++ )  
		if( numb[i] ){
			c += max(0LL,1LL*x*numb[i]-numa[i]);
		}
		return c <= num;
	}
	int main(){
		num = 0;
		gets(a);
		gets(b);
		lena = strlen(a);
		lenb = strlen(b);
		for( int i = 0 ; i < lena ; i++ ){
			if( a[i] >= 'a' && a[i] <= 'z' )
				numa[a[i]-'a'] ++;
			else num ++ ;
		}
		for( int i = 0 ; i < lenb ; i++ ){
			numb[b[i]-'a'] ++;
		}
		int  l = 0 , r = AX , x = 0;
		while( l <= r ){
			int mid = (l+r)>>1;
			if( check(mid) ) { l = mid + 1 ; x = mid ; }
			else r = mid - 1;
		}
		int id = 0;
		for( int i = 0 ; i < lena ; i++ ){         
			if( a[i] != '?' ) printf("%c",a[i]);
			else{
				while( id < 26 && numa[id] >= numb[id]*x ) id++;
				if( id < 26 ){
					printf("%c", id + 'a');
					numa[id]++;
				}else{
					putchar('a');
				}
			}
		}
		return 0;
	}