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
 #include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int a[+];
char temp[+];
bool check(const char t[],const char p[],int mid)
{
memset(temp,,sizeof(temp));
strcpy(temp,t); //拷贝到一个临时数组
for(int i=;i<=mid;i++) temp[a[i]-]=''; //把要划掉的那几个字母依次划掉
for(int i=,j=,temp_len=strlen(t),p_len=strlen(p);i<temp_len;i++){
if(temp[i] == p[j]) j++;
if(j == p_len) return true;
}
return false;
}
int main()
{
char t[+],p[+];
scanf("%s%s",t,p);
int t_len=strlen(t);
for(int i=;i<=t_len;i++) scanf("%d",&a[i]);
int st=,ed=t_len;
while(ed-st>)
{
int mid=st+(ed-st)/;
if(check(t,p,mid)) st=mid;
else ed=mid;
}
printf("%d\n",st);
}

开始的时候一直在test 10上RUNTIME_ERROR,一直找不出原因,刚开始以为二分那边有问题,换了网上AC的二分方式依然是这样,后来才发现开int数组a[]的大小时候

200000+5写成了20000+5……真是僵硬……看来预定义一个MAXN不是没有道理的……

codeforces 779D - String Game的更多相关文章

  1. CodeForces 779D. String Game(二分答案)

    题目链接:http://codeforces.com/problemset/problem/779/D 题意:有两个字符串一个初始串一个目标串,有t次机会删除初始串的字符问最多操作几次后刚好凑不成目标 ...

  2. CodeForces - 779D String Game 常规二分

    题意:给你两个串,S2是S1 的一个子串(可以不连续).给你一个s1字符下标的一个排列,按照这个数列删数,问你最多删到第几个时S2仍是S1 的一个子串. 题解:二分删掉的数.判定函数很好写和单调性也可 ...

  3. CodeForces - 779D String Game(二分)

    Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But i ...

  4. 【codeforces 779D】String Game

    [题目链接]:http://codeforces.com/contest/779/problem/D [题意] 给你一段操作序列; 按顺序依次删掉字符串1中相应位置的字符; 问你最多能按顺序删掉多少个 ...

  5. Codeforces 799D. String Game 二分

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

  6. 779D. String Game 二分 水

    Link 题意: 给出两字符串$a$,$b$及一个序列,要求从前往后按照序列删掉$a$上的字符,问最少删多少使$b$串不为a的子串 思路: 限制低,直接二分答案,即二分序列位置,不断check即可. ...

  7. Codeforces - 828C String Reconstruction —— 并查集find()函数

    题目链接:http://codeforces.com/contest/828/problem/C C. String Reconstruction time limit per test 2 seco ...

  8. CodeForces 159c String Manipulation 1.0

    String Manipulation 1.0 Time Limit: 3000ms Memory Limit: 262144KB This problem will be judged on Cod ...

  9. Codeforces 710F String Set Quries

    题意 维护一个字符串的集合\(D\), 支持3种操作: 插入一个字符串\(s\) 删除一个字符串\(s\) 查询一个字符串\(s\)在\(D\)中作为子串出现的次数 强制在线 解法 AC自动机+二进制 ...

随机推荐

  1. RHEL5 yum更新源

    1.检查yum是否安装 rpm -qa |grep yum 2.利用CentOS的yum更新源来实现RHEL5的YUM功能 vi /etc/yum.repos.d/CentOS-Base.repo [ ...

  2. HDU 5083 Instruction(字符串处理)

    Problem Description Nowadays, Jim Green has produced a kind of computer called JG. In his computer, ...

  3. C语言编程规范—命名规则

    C是一门朴素的语言,你使用的命名也应该这样.与Modula-2和Pascal程序员不同,C程序员不使用诸如“ThisVariableIsATemporaryCounter”这样“聪明”的名字.C程序员 ...

  4. 【Android】水平居中 垂直居中 中心居中

    android:layout_centerInParent 将该组件放置于水平方向中央及垂直中央的位置 android:layout_centerHorizontal 将该组件放置于水平方向中央的位置 ...

  5. kohana 简单使用

    声明:基于公司使用的 Kohana 框架写的,不确定是否适用于原生 Kohana 附:Kohana 3 中文手册,传送门:http://www.lampblog.net/kohana3%E4%BD%B ...

  6. java的代理和动态代理简单测试

    什么叫代理与动态代理? 1.以买火车票多的生活实例说明. 因为天天调bug所以我没有时间去火车票,然后就给火车票代理商打电话订票,然后代理商就去火车站给我买票.就这么理解,需要我做的事情,代理商帮我办 ...

  7. Kubernetes 集群:规划与搭建

    Kubernetes 集群环境: IP地址 主机名 角色 软硬件限制 192.168.119.134 master1 deploy ,master1 ,lb1 ,etcd (1) CPU至少1核,内存 ...

  8. 使用 urllib 构造请求对象

    (1) urllib.request.urlopen()方法可以实现最基本请求的发起,但这几个简单的参数并不足以构建一个完整的请求(2) 我们可以使用 urllib.request.Request() ...

  9. springboot 集成elasticsearch5.4.3

    官网上对elasticsearch 的集成用的是spring-data,而且,暂时不支持5.x的版本, 要是想集成5.x的版本,我们只能在pom.xml文件中进行修改,如图: <project ...

  10. vmp3.0.9全保护拆分解析

    https://mp.weixin.qq.com/s/WO6w_L-cYwH5KB2rilZdag 以下为了避免插件干扰,故采用x64dbg原版进行分析. 首先我通过检测到调试器的弹窗进行栈回溯,定位 ...