CodeForces 762C Two strings

今天补一下前天的一道题，我果然是太菜了TAT

题面如下：

C. Two strings

You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.

Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.

Input

The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 10⁵ characters.

Output

On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.

If the answer consists of zero characters, output «-» (a minus sign).

Examples

Input

hi
bob

Output

-

Input

abca
accepted

Output

ac

Input

abacaba
abcdcba

Output

abcba

Note

In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.

In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b.

题意：给你两个字符串A，B，让你从B中删除最短的一段连续字符，使得B成为A的子序列。无法办到就输出'-'。

分析：日常暴力一发。枚举左右两个需要删除的区间，并且对每一次剩下来的字符串作匹配，看是否是A的子序列。枚举区间需要O(n^2)，匹配需要O(n)，复杂度是不能通过的。

这时候我猛然就想起了敬爱的雨神的谆谆教诲：“求最x值最x，不是贪心就是二分。”emmm

这道题的话贪心肯定不靠谱，于是就只能二分了，二分需要删除的区间长度。为了实现这个二分的操作，我们需要先对这两个字符串进行预处理。我们开两个数组pos1,pos2，pos1[i],pos2[i]分别代表字符串B从前往后第i个字符在A串里第一次出现的位置和从后往前倒数第i个字符在A串里第一次出现的位置。于是匹配的时候我们就先judge一手，如果发现对于某个答案mid而言，把left赋为0，right赋为mid+1，若pos1[left]

据说还有一种做法是用双指针扫描，在B字符串中枚举left，向右找到right使得pos1[left]

代码贴下来：

#include
#include
#include
#include
using namespace std;

const int maxn=1e6+7;
int ans,ansl,ansr,alen,blen;
int pos1[maxn],pos2[maxn];
char a[maxn],b[maxn];

void prework(){//预处理两个pos数组的过程
    memset(pos1,0,sizeof(pos1));
    memset(pos2,0,sizeof(pos2));
    alen=strlen(a+1),blen=strlen(b+1);
    int l=1;
    for(int i=1;i<=blen;i++){
        while(l<=alen&&a[l]!=b[i]) l++;
        pos1[i]=l;
        if(l<=alen)
            l++;
    }
    pos1[0]=0;
    pos2[blen+1]=alen+1;
    int r=alen+1;
    for(int i=blen;i>=1;i--){
        while(l>=1&&a[l]!=b[i]) l--;
        pos2[i]=l;
        if(l>=1)
            l--;
    }
}

bool jug(int mid){//判断当前答案mid是否合法
    int l=0,r=mid+1;
    while(r<=blen+1){
        if(pos1[l]>1;
        if(jug(mid))
            r=mid;
        else
            l=mid;
    }
    if(jug(l)) ans=l;//判断l和r哪一个是答案
    else if(jug(r)) ans=r;
    else ans=r+1;
    l=0,r=ans+1;
    ansl=0;
    ansr=blen+1;
    while(r<=blen+1){//分别找到删除区间的左右边界在哪
        if(pos1[l]=1;i--)
//            cout<