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

B. Pasha and String

time limit per test

memory limit per test

input

output

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|

m days performing the following transformations on his string — each day he chose integer ai and reversed a piece of string (a segment) from position ai to position|s| - ai. It is guaranteed that 2·ai ≤ |s|.

m

Input

s of length from 2 to 2·105

m (1 ≤ m ≤ 105) —  the number of days when Pasha changed his string.

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

Output

s will look like after m

Sample test(s)

input

abcdef 1 2

output

aedcbf

input

vwxyz 2 2 2

output

vwxyz

input

abcdef 3 1 2 3

output

fbdcea

     思路：因为按照题意所有的翻转都是按照中心对称翻，也就是说可以把翻转的次数记录下来，如果能被2整除就说明通过翻转他又回到了原来的地方，如果不能被整除就翻转一次。最后输出字符串。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int a[210000];
char str[210000];
int n,m;

int main()
{
    while(scanf("%s",str)!=EOF)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        int x;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            a[x-1]++;
        }
        int l = strlen(str);
        for(int i=1;i<=l/2;i++)
        {
            a[i] += a[i-1];
        }
        for(int i=0;i<l/2;i++)
        {
            if(a[i]%2 != 0)
            {
                char t = str[i];
                str[i] = str[l-1-i];
                str[l-1-i] = t;
            }
        }
        printf("%s\n",str);
    }
    return 0;
}