Codeforces Round #297 (Div. 2)B. Pasha and String 前缀和

2021-09-08 22:33:41

Codeforces Round #297 (Div. 2)B. Pasha and String

Time Limit: 2 Sec Memory Limit: 256 MB
Submit: xxx Solved: 2xx

题目连接

http://codeforces.com/contest/525/problem/B

Description

Pasha got a very beautiful string s for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |s| from left to right, where |s| is the length of the given string.

Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent m days performing the following transformations on his string — each day he chose integer a_i and reversed a piece of string (a segment) from position a_i to position |s| - a_i + 1. It is guaranteed that 2·a_i ≤ |s|.

You face the following task: determine what Pasha's string will look like after m days.

Input

The first line of the input contains Pasha's string s of length from 2 to 2·10⁵ characters, consisting of lowercase Latin letters.

The second line contains a single integer m (1 ≤ m ≤ 10⁵) — the number of days when Pasha changed his string.

The third line contains m space-separated elements a_i (1 ≤ a_i; 2·a_i ≤ |s|) — the position from which Pasha started transforming the string on the i-th day.

Output

In the first line of the output print what Pasha's string s will look like after m days.

Sample Input

Input

abcdef
1
2

Input

vwxyz
2
2 2

Input

abcdef
3
1 2 3

Sample Output

Output

aedcbf

Output

vwxyz

Output

fbdcea

HINT

题意：

给你一个字符串，然后给你一个数字i，然后i到s.size()-i直接的位置全部翻转，然后问你m次操作之后，是什么样子

题解：

首先，只有翻转奇数次和翻转偶数次两种情况，我们用一个类似前缀和的操作，就可以处理出这个位置究竟翻转了多少次，然后输出就好啦

~\(≧▽≦)/~啦啦啦，这道题完啦

代码：

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 400010

#define mod 10007

#define eps 1e-9

//const int inf=0x7fffffff;   //无限大

const int inf=0x3f3f3f3f;

/*

*/

//**************************************************************************************

inline ll read()

{

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int flag[maxn];

int main()

{

    string s;

    cin>>s;

    int n;

    cin>>n;

    for(int i=;i<n;i++)

    {

        int x=read();

        flag[x-]++;

    }

    for(int i=;i<=s.size()/;i++)

    {

        flag[i]+=flag[i-];

    }

    for(int i=;i<s.size();i++)

    {

        if(i<s.size()/)

        {

            if(flag[i]%==)

                cout<<s[i];

            else

                cout<<s[s.size()-i-];

        }

        else

        {

            if(flag[s.size()-i-]%==)

                cout<<s[i];

            else

                cout<<s[s.size()-i-];

        }

    }

}

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