Codeforces--525B--Pasha and String（思维）



Pasha and String

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

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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 integera_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

abcdef12

Output

aedcbf

Input

vwxyz22 2

Output

vwxyz

Input

abcdef31 2 3

Output

fbdcea

题意：给定一个字符串，然后给出m个位置，x表示x--l-x+1之间的的字符全部翻转，输出最后产生的字符串

就是一个水题，可以用一个数组记录每一位的翻转，累计的时候通过奇偶来判断翻转的次数，奇数就翻转

不知道为什么，！=EOF就是不过

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char s[20000000+100];int num[2000000+100];int main(){while(~scanf("%s",s+1)){int l=strlen(s+1);memset(num,0,sizeof(num));int m,x;scanf("%d",&m);for(int i=0;i<m;i++){scanf("%d",&x);num[x]++;}int ans=0;for(int i=1;i<=l/2;i++){ans+=num[i];if(ans%2==1)swap(s[i],s[l-i+1]);}printf("%s\n",s+1);}return 0;}