hust新人赛模拟20150407 F

F - F

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

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

Output

aedcbf

Input

vwxyz
2
2 2

Output

vwxyz

Input

abcdef
3
1 2 3

Output

fbdcea

题意是说一个字符串，进行m次颠倒变换（从a[i]位置到a[l-i+1]位置），问得到的字符串。
容易发现，对于越在里边（对称，也就是越靠近中间位置）的字符，调换的次数越多。
我们可以把a[i]从小到大排序。
然后经过分析发现，把两个相邻的a[i]分为一组，做处理，如果m为奇数，最后还剩下a[m]没有被分组，要单独处理a[m]
细节上要注意st数组是从st[0]开始的...好吧的确不方便，适牛也说我了。。数组下标以后还是从0开始吧。。。主要是受高中OI用的pascal的影响。。。那个数组下标随便啊。

代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>

using namespace std;

int m,k,len;
const int N=2E5+7;
int a[N];
char st[N];

int main()
{
    cin>>st;
    scanf("%d",&m);
    for ( int i = 1 ; i <= m ; i++ )
        scanf("%d",&a[i]);
    sort(a+1,a+m+1);
    k = 1;
    len = strlen(st);
    while (k<=m)
    {
        for ( int j = a[k] ; j <= a[k+1]-1 ; j++)
            swap(st[j-1],st[len-j]);
        k = k + 2;
    }
    if ( m %2==1 )
        for ( int i = a[m]; i <= len/2 ; i++ )
            swap(st[i-1],st[len-i]);
    cout<<st<<endl;
    return 0;
}