D. 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 ta1... 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
ababcba
abb
5 3 4 1 7 6 2
output
3
input
bbbabb
bb
1 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.

 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
#include<stack>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
char s1[200003],s2[200003];
int del[200003];
int vis[200003];
 
bool judge(int t, int s1_len, int s2_len)
{
    for(int i=0; i<s1_len; i++)
    {
        vis[i] = 0;
    }
    for(int i=0; i<t; i++)
    {
        vis[del[i] - 1] = 1;
    }
    int j = 0;
    for(int i=0; i<s1_len; i++)
    {
        if(vis[i]==0 && s1[i] == s2[j]) j++;
        if( j >= s2_len) return true;
    }
    return false;
}
 
int main()
{
    scanf("%s",s1);
    scanf("%s",s2);
    int length1 = strlen(s1);
    int length2 = strlen(s2);
    for(int i=0; i<length1; i++)
    {
        scanf("%d",&del[i]);
    }
    int l = 0;
    int r = length1 - 1;
    int ans;
    while(l <= r)
    {
        int mid = (l + r) >> 1;
        if(judge(mid, length1, length2)){
            l = mid + 1;
            ans = mid;
        }
        else{
            r = mid - 1;
        }
    }
    printf("%d",ans);
}

Codeforces Round #402 (Div. 2) D. String Game的更多相关文章

  1. Codeforces Round #402 (Div. 2) D. String Game(二分答案水题)

    D. String Game time limit per test 2 seconds memory limit per test 512 megabytes input standard inpu ...

  2. Codeforces Round #402 (Div. 2) D String Game —— 二分法

    D. String Game time limit per test 2 seconds memory limit per test 512 megabytes input standard inpu ...

  3. 【二分答案】Codeforces Round #402 (Div. 2) D. String Game

    二分要删除几个,然后暴力判定. #include<cstdio> #include<cstring> using namespace std; int a[200010],n, ...

  4. Codeforces Round #402 (Div. 2) A+B+C+D

    Codeforces Round #402 (Div. 2) A. Pupils Redistribution 模拟大法好.两个数列分别含有n个数x(1<=x<=5) .现在要求交换一些数 ...

  5. Codeforces Round #402 (Div. 2)

    Codeforces Round #402 (Div. 2) A. 日常沙比提 #include<iostream> #include<cstdio> #include< ...

  6. Codeforces Round #402 (Div. 2) A,B,C,D,E

    A. Pupils Redistribution time limit per test 1 second memory limit per test 256 megabytes input stan ...

  7. Codeforces Round #402 (Div. 2) A B C sort D二分 (水)

    A. Pupils Redistribution time limit per test 1 second memory limit per test 256 megabytes input stan ...

  8. Codeforces Round #402 (Div. 2) 题解

    Problem A: 题目大意: 给定两个数列\(a,b\),一次操作可以交换分别\(a,b\)数列中的任意一对数.求最少的交换次数使得任意一个数都在两个序列中出现相同的次数. (\(1 \leq a ...

  9. CodeForces Round #402 (Div.2) A-E

    2017.2.26 CF D2 402 这次状态还算能忍吧……一路不紧不慢切了前ABC(不紧不慢已经是在作死了),卡在D,然后跑去看E和F——卧槽怎么还有F,早知道前面做快点了…… F看了看,不会,弃 ...

随机推荐

  1. stark组件之展示数据(查)

      1.编辑按钮构建完成 2.构造表头,删除,checkbox,links编辑 3.代码+总结   1.编辑按钮构建完成 1.必备知识预习 第一个会打印5. 第二个输出alex alex是person ...

  2. 学习yii2.0——依赖注入

    依赖注入 依赖注入是一种设计模式,可以搜索“php依赖注入”,这里不阐述了. yii框架的依赖注入 Yii 通过 yii\di\Container 类提供 DI 容器特性. 它支持如下几种类型的依赖注 ...

  3. Hive简单编程实践-词频统计

    一.使用MapReduce的方式进行词频统计 (1)在HDFS用户目录下创建input文件夹 hdfs dfs -mkdir input 注意:林子雨老师的博客(http://dblab.xmu.ed ...

  4. Apache Tomcat® - Which Version Do I Want?

    Apache Tomcat® - Which Version Do I Want?http://tomcat.apache.org/whichversion.html

  5. Python3练习题 006 冒泡排序

    import random a = [random.randint(1,100) for i in range(10)]def bu(target): length = len(target) whi ...

  6. spark、standalone集群 (2)集群zookeeper 热备

     测试 cmd     spark-examples-1.6.0-hadoop2.6.0.jar   spark 2.0以后  就没有这个 jar.需要下载 ./bin/spark-submit -- ...

  7. Sqlserver tablediff的简单使用

    1. 先列举一下自己简单的比较语句 tablediff -sourceserver 10.24.160.73 -sourcedatabase cwbasemi70 -sourceschema lcmi ...

  8. jquery获取select多选框选中的值

    select下拉框选中的值,用jquery大家应该都会获取, $("#selectBox option:selected").val(); 如果select是多选的,也这么获取的话 ...

  9. php的amqp扩展 安装(windows) rabbitmq学习篇

    因为RabbitMQ是由erlang语言实现的,所以先要安装erlang环境erlang 下载安装 http://www.erlang.org/download.htmlrabbitmq 下载安装 h ...

  10. VMware与CentOS的安装与Linux简单指令

    一 . VMware与CentOS系统安装 下载CentOS系统的ISO镜像 # 官方网站,国外网站,下载速度会很慢 www.centos.org # 由于国外的下载速度慢,我们可以使用国内的镜像源 ...