ACM刷题之Codeforces ————String Game

 String Game

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.

Sergey gives Nastya the word t and wants to get the word p out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters' indices of the word t: a1... a|t|. We denote the length of word x as |x|. Note that after removing one letter, the indices of other letters don't change. For example, if t = "nastya" and a = [4, 1, 5, 3, 2, 6] then removals make the following sequence of words "nastya"  "nastya"  "nastya"  "nastya"  "nastya"  "nastya"  "nastya".

Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word p. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.

It is guaranteed that the word p can be obtained by removing the letters from word t.

Input

The first and second lines of the input contain the words t and p, respectively. Words are composed of lowercase letters of the Latin alphabet (1 ≤ |p| < |t| ≤ 200 000). It is guaranteed that the word p can be obtained by removing the letters from word t.

Next line contains a permutation a1, a2, ..., a|t| of letter indices that specifies the order in which Nastya removes letters of t (1 ≤ ai ≤ |t|, all ai are distinct).

Output

Print a single integer number, the maximum number of letters that Nastya can remove.

Examples

input

ababcbaabb5 3 4 1 7 6 2

output

3

input

bbbabbbb1 6 3 4 2 5

output

4

Note

In the first sample test sequence of removing made by Nastya looks like this:

"ababcba"  "ababcba"  "ababcba"  "ababcba"

Nastya can not continue, because it is impossible to get word "abb" from word "ababcba".

So, Nastya will remove only three letters.

在比赛的时候这题没做出来。

因为感觉数据量太大，直接暴力的话会超时。

赛后看了下大神们的代码，发现是二分。

自己研究了下，中间错了好多次。

下面总结下这里的二分。

while(l<=r)
{
if(mids(MID(l,r)))
{
l=MID(l,r)+1;
}else{
r=MID(l,r)-1;
}
}

这里是二分的关键代码，注意每次都要 l+1  和 r-1 否则会出现死循环。

注意，最后输出答案的时候，l要减1,因为最后一次出while循环l会额外加1.

下面是ac代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>#include<map>#include<set>#include<queue>#include<string>#include<iostream>using namespace std;#define MID(x,y) ((x+y)>>1)#define CLR(arr,val) memset(arr,val,sizeof(arr))#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const int N=2e5+7;int l,r,n,len2,r2;char b[200002];char a[200002];bool c[200002];int d[200002];bool mids(int mi){CLR(c,false);for(int i=1;i<=mi;i++){c[d[i]]=true;}for(int i=1,j=1;i<=r2;i++){if((!c[i])&&(a[i]==b[j])){++j;if(j>len2){return true;}}}return false;}int main(){//freopen("f:/input.txt", "r", stdin);int len1;while(scanf("%s%s",a+1,b+1)!=EOF){CLR(d,0);r=strlen(a+1);for(int i=1;i<=r;i++)scanf("%d",&d[i]);len2=strlen(b+1);r2=r;l=0;while(l<=r){if(mids(MID(l,r))){l=MID(l,r)+1;}else{r=MID(l,r)-1;}}printf("%d\n",l-1);} }