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

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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int flag[maxn];
int main()
{
    string s;
    cin>>s;
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        int x=read();
        flag[x-1]++;
    }
    for(int i=1;i<=s.size()/2;i++)
    {
        flag[i]+=flag[i-1];
    }
    for(int i=0;i<s.size();i++)
    {
        if(i<s.size()/2)
        {
            if(flag[i]%2==0)
                cout<<s[i];
            else
                cout<<s[s.size()-i-1];
        }
        else
        {
            if(flag[s.size()-i-1]%2==0)
                cout<<s[i];
            else
                cout<<s[s.size()-i-1];
        }
    }
}